KC Sinha: Exercise 4.1 – Mathematics Solution Class 10 Chapter 4 Real numbers

Question 1

From the given figure, find the value of the following:

(i) sin C
(ii) sin A
(iii) cos C
(iv) cos A
(v) tan C
(vi) tan A
Sol :
(i) Sin C
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
Side opposite to ∠C = AB = 3
Hypotenuse = AC = 5
So, $\sin C=\frac{A B}{A C}=\frac{3}{5}$

(ii) Sin A
So, here θ = A
The side opposite to ∠A = BC = 4
Hypotenuse = AC = 5
So, $\sin A=\frac{B C}{A C}=\frac{4}{5}$

(iii) Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
Side adjacent to ∠C = BC = 4
Hypotenuse = AC = 5
So, $\cos C=\frac{B C}{A C}=\frac{4}{5}$

(iv) Cos A
Here, θ = A
Side adjacent to ∠A = AB = 3
Hypotenuse = AC = 5
So, $\cos A=\frac{A B}{A C}=\frac{3}{5}$

(v) tan C
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
So, here θ = C
Side opposite to ∠C = AB = 3
Side adjacent to ∠C = BC = 4
So, $\tan C=\frac{A B}{B C}=\frac{3}{4}$

(vi) tan A
here θ = A
Side opposite to ∠A = BC = 4
Side adjacent to ∠A = AB = 3
So, $\tan A=\frac{A B}{B C}=\frac{4}{3}$

Question 2 

From the given figure, find the value of :

(i) tan θ
(ii) cos θ

Sol :
(i) tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to θ = AB = 4
Side adjacent to θ = BC = 3
So, $\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{4}{3}$

(ii) cos θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to θ = BC = 3
Hypotenuse = AC = 5
So, $\cos \theta=\frac{B C}{A C}=\frac{3}{5}$

Question 3 

From the given figure, find the value of

(i) sin θ
(ii) tan θ
(iii) tan A – cot C

Sol :
(i) sin θ
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to θ = BC = ?
Hypotenuse = AC = 13
Firstly we have to find the value of BC.
So, we can find the value of BC with the help of Pythagoras theorem.
According to Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
⇒ (AB)² + (BC)² = (AC)²
⇒ (12)² + (BC)² = (13)²
⇒ 144 + (BC)² = 169
⇒ (BC)² = 169–144
⇒ (BC)² = 25
⇒ BC =√25
⇒ BC =±5
But side BC can’t be negative. So, BC = 5
Now, BC = 5 and AC = 13
So, $\sin \theta=\frac{B C}{A C}=\frac{5}{13}$

(ii) tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to θ = BC = 5
Side adjacent to θ = AB = 12
So, $\tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$

(iii) tan A – cot C
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
and
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
tan A
Here, θ = A
Side opposite to ∠A = BC = 5
Side adjacent to ∠A = AB = 12
So, $\tan \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$
Cot C
Here, θ = C
Side adjacent to ∠C = BC = 5
Side opposite to ∠C = AB = 12
So, $\cot C=\frac{B C}{A B}=\frac{5}{12}$
So, $\tan A-\cot C=\frac{5}{12}-\frac{5}{12}=0$

Question 4 A 

In ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine

a. sin A, cos A
b. sin C, cos C
Sol :

(i)
(a) sin A
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
So, here θ = A
Side opposite to ∠A = BC = 7
Hypotenuse = AC = ?
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem.

According to Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
⇒ (AB)² + (BC)² = (AC)²
⇒ (24)² + (7)² = (AC)²
⇒ 576 + 49 = (AC)²
⇒ (AC)² = 625
⇒ AC =√625
⇒ AC =±25
But side AC can’t be negative. So, AC = 25cm
Now, BC = 7 and AC = 25
So, $\sin A=\frac{B C}{A C}=\frac{7}{25}$
Cos A
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
So, here θ = A
Side adjacent to ∠A = AB = 24
Hypotenuse = AC = 25
So,$\cos A=\frac{A B}{A C}=\frac{24}{25}$
(b) sin C

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
The side opposite to ∠C = AB = 24
Hypotenuse = AC = 25
So, $\sin C=\frac{A B}{A C}=\frac{24}{25}$
Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
Side adjacent to ∠C = BC = 7
Hypotenuse = AC = 25
So, $\cos C=\frac{B C}{A C}=\frac{7}{25}$

Question 4 B 

Consider ∆ACB, right angled at C, in which AB = 29 units, BC = 21 units and ∠ABC=θ. Determine the values ofa. cos² θ+ sin² θ
b. cos² θ – sin² θ
Sol :(a) Cos²θ +sin² θ
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem.
According to Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
⇒ (AC)² + (BC)² = (AB)²
⇒ (AC)² + (21)² = (29)²
⇒ (AC)² = (29)² – (21)²
Using the identity a² –b² = (a+b) (a – b)
⇒ (AC)² = (29–21)(29+21)
⇒ (AC)² = (8)(50)
⇒ (AC)² = 400
⇒ AC =√400
⇒ AC =±20
But side AC can’t be negative. So, AC = 20units
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
In ∆ACB, Side opposite to angle θ = AC = 20
and Hypotenuse = AB = 29
So, $\sin \theta=\frac{A C}{A B}=\frac{20}{29}$
Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$

In ∆ACB, Side adjacent to angle θ = BC = 21
and Hypotenuse = AB = 29
So, $\cos \theta=\frac{B C}{A B}=\frac{21}{29}$
So$\cos ^{2} \theta+\sin ^{2} \theta=\left(\frac{21}{29}\right)^{2}+\left(\frac{20}{29}\right)^{2}$
$=\frac{441+400}{29 \times 29}$
$=\frac{841}{841}$
=1
Cos²θ +sin² θ = 1
(b) Cos²θ – sin² θ
Putting values, we get
$\cos ^{2} \theta-\sin ^{2} \theta=\left(\frac{21}{29}\right)^{2}-\left(\frac{20}{29}\right)^{2}$
$=\frac{441-400}{29 \times 29}$
$=\frac{41}{841}$

Question 4 C 

In ∆ABC, ∠A is a right angle, then find the values of sin B, cos C and tan B in each of the following :a. AB = 12, AC = 5, BC = 13
b. AB = 20, AC = 21, BC = 29
c. BC = √2, AB = AC = 1
Sol :
Given that ∠A is a right angle.(a) AB = 12, AC = 5, BC = 13
To Find : sin B, cos C and tan B
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = B
Side opposite to angle B = AC = 5
Hypotenuse = BC =13
So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{5}{13}$
Now, Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = C
Side adjacent to angle C = AC = 5
Hypotenuse = BC =13
So, $\cos C=\frac{A C}{B C}=\frac{5}{13}$
Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = B
The side opposite to angle B = AC = 5
The side adjacent to angle B = AB = 12
So, $\tan \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{5}{12}$
(b) AB = 20, AC = 21, BC = 29
To Find: sin B, cos C and tan B
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = B
The side opposite to angle B = AC =21
Hypotenuse = BC =29
So, $\sin B=\frac{A C}{B C}=\frac{21}{29}$
Now, Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = C
Side adjacent to angle C = AC = 21
Hypotenuse = BC = 29
So,$\cos C=\frac{A C}{B C}=\frac{21}{29}$

Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = B
The side opposite to angle B = AC = 21
The side adjacent to angle B = AB = 20
So, $\tan \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{21}{20}$

(c) BC =√2, AB = AC = 1
To Find: sin B, cos C and tan B
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = B
The side opposite to angle B = AC =1
Hypotenuse = BC =√2
So$\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{1}{\sqrt{2}}$
Now, Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = C
Side adjacent to angle C = AC = 1
Hypotenuse = BC = √2
So, $\cos C=\frac{A C}{B C}=\frac{1}{\sqrt{2}}$
Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = B
The side opposite to angle B = AC = 1
The side adjacent to angle B = AB = 1
So,$\tan B=\frac{A C}{A B}=\frac{1}{1}=1$

Question 5 A 

Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below :

Sol :
Firstly, we give the name to the midpoint of BC i.e. M
BC = BM + MC = 2BM or 2MC
⇒ BM = 5 and MC = 5
Now, we have to find the value of AM, and we can find out with the help of Pythagoras theorem.
So, In ∆AMB
⇒ (AM)² + (BM)² = (AB)²
⇒ (AM)² + (5)² = (13)²
⇒ (AM)² = (13)² – (5)²
Using the identity a² –b² = (a+b) (a – b)
⇒ (AM)² = (13–5)(13+5)
⇒ (AM)² = (8)(18)
⇒ (AM)² = 144
⇒ AM =√144
⇒ AM =±12
But side AM can’t be negative. So, AM = 12
a. sin θ
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
In ∆AMB
Side opposite to θ = AM = 12
Hypotenuse = AB=13
So, $\sin \theta=\frac{A M}{A B}=\frac{12}{13}$
So, $\sin \theta=\frac{12}{13}$
b. cos θ
We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
In ∆AMB
The side adjacent to θ = BM = 5
Hypotenuse = AB = 13
So, $\cos \theta=\frac{B M}{A B}=\frac{5}{13}$
So, $\cos \theta=\frac{5}{13}$
c. tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
In ∆AMB
Side opposite to θ = AM = 12
The side adjacent to θ = BM = 5
So, $\tan \theta=\frac{\mathrm{AM}}{\mathrm{BM}}=\frac{12}{5}$
So, $\tan \theta=\frac{12}{5}$

Question 5 B 

Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below :
Sol :
Firstly, we have to find the value of XM and we can find out with the help of Pythagoras theorem
So, In ∆XMZ
⇒ (XM)² + (MZ)² = (XZ)²
⇒ (XM)² + (16)² = (20)²
⇒ (XM)² = (20)² – (16)²
Using the identity a² –b² = (a+b) (a – b)
⇒ (XM)² = (20–16)(20+16)
⇒ (XM)² = (4)(36)
⇒ (XM)² = 144
⇒ XM =√144
⇒ XM =±12
But side XM can’t be negative. So, XM = 12
Now, In ∆XMY we have the value of XM and MY but we don’t have the value of XY.
So, again we apply the Pythagoras theorem in ∆XMY
⇒ (XM)² + (MY)² = (XY)²
⇒ (12)² + (5)² = (XY)²
⇒ 144 + 25 = (XY)²
⇒ (XY)² = 169
⇒ XY =√169
⇒ XY =±13
But side XY can’t be negative. So, XY = 13
a. sin θ
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
In ∆XMY
Side opposite to θ = MY = 5
Hypotenuse = XY = 13
So, $\sin \theta=\frac{\mathrm{MY}}{\mathrm{XY}}=\frac{5}{13}$
b. cos θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
In ∆XMY
Side adjacent to θ = XM = 12
Hypotenuse = XY = 13
So,$\cos \theta=\frac{\mathrm{XM}}{\mathrm{XY}}=\frac{12}{13}$
c. tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
In ∆XMY
The side opposite to θ = MY = 5
Side adjacent to θ = XM = 12
So, $\tan \theta=\frac{\mathrm{MY}}{\mathrm{XM}}=\frac{5}{12}$

Question 6 

In ∆PQR, ∠Q is a right angle PQ = 3, QR = 4. If ∠P=α and ∠R=β, then find the values of(i) sin α (ii) cos α
(iii) tan α (iv) sin β
(v) cos β (vi) tan β
Sol :
Given : PQ = 3, QR = 4
⇒ (PQ)² + (QR)² = (PR)²
⇒ (3)² + (4)² = (PR)²
⇒ 9 + 16 = (PR)²
⇒ (PR)² = 25
⇒ PR =√25
⇒ PR =±5
But side PR can’t be negative. So, PR = 5
(i) sin α
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = α
The side opposite to angle α = QR =4
Hypotenuse = PR =5
So, $\sin \alpha=\frac{4}{5}$
(ii) cos α
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = α
The side adjacent to angle α = PQ =3
Hypotenuse = PR =5
So, $\cos \alpha=\frac{3}{5}$
(iii) tan α
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = α
Side opposite to angle α = QR =4
Side adjacent to angle α = PQ =3
So,$\tan \alpha=\frac{4}{3}$
(iv) sin β
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = β
The side opposite to angle β = PQ =3
Hypotenuse = PR =5
So, $\sin \beta=\frac{3}{5}$
(v) cos β
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = β
Side adjacent to angle β = QR =4
Hypotenuse = PR =5
So, $\cos \beta=\frac{4}{5}$
(vi) tan β
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = β
Side opposite to angle β = PQ =3
Side adjacent to angle β = QR =4
So, $\tan \beta=\frac{3}{4}$

Question 7 A 

If $\sin \theta=\frac{4}{5}$ ,then find the values of cos θ and tan θ.Sol :
Given: $\sin \theta=\frac{4}{5}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{5}$
Let,
Perpendicular =AB =4k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (BC)² = (5k)²
⇒ 16k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –16 k²
⇒ (BC)² = 9 k²
⇒ BC =√9 k²
⇒ BC =±3k
But side BC can’t be negative. So, BC = 3k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
The side adjacent to angle θ or base = BC =3k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$
Now,
We know that,$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =4k
Base = BC =3k
So, $\tan \theta=\frac{4 k}{3 k}=\frac{4}{3}$

Question 7 B 

If $\sin \mathrm{A}=\frac{3}{4}$ ,calculate cos A and tan A.Sol :
Given: Sin A $=\frac{3}{4}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{4} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}$
Let,
Side opposite to angle θ = BC =3k
and Hypotenuse = AC =4k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (3k)² = (4k)²
⇒ (AB)² + 9k² = 16k²
⇒ (AB)² = 16 k² – 9 k²
⇒ (AB)² = 7 k²
⇒ AB =k√7
So, AB = k√7
Now, we have to find the value of cos A and tan A
We know that,
$\cos \theta=\frac{\text { Side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = A
The side adjacent to angle A = AB =k√7
Hypotenuse = AC =4k
So,$\cos A=\frac{k \sqrt{7}}{4 k}=\frac{\sqrt{7}}{4}$
Now,
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
The side opposite to angle A = BC =3k
The side adjacent to angle A = AB =k√7
So,$\tan \mathrm{A}=\frac{3 \mathrm{k}}{\mathrm{k} \sqrt{7}}=\frac{3}{\sqrt{7}}$

Question 8 

If $\sin \theta=\frac{3}{5}$ , then find the values cos θ and tan θ.Sol :
Given:$\sin \theta=\frac{3}{5}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$
Let,
Perpendicular =AB =3k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (BC)² = (5k)²
⇒ 9k² + (BC)² = 25k²
⇒ (BC)² = 25 k² – 9 k²
⇒ (BC)² = 16 k²
⇒ BC =√16 k²
⇒ BC =±4k
But side BC can’t be negative. So, BC = 4k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = BC =4k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{4 \mathrm{k}}{5 \mathrm{k}}=\frac{4}{5}$
Now, tan θ
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =3k
Base = BC =4k
So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$

Question 9 

If $\cos \theta=\frac{4}{5}$ , then find the value of tan θ.Sol :
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$
$\cos \theta=\frac{4}{5} \Rightarrow \frac{B}{H}=\frac{4}{5} \Rightarrow \frac{B C}{A C}=\frac{4}{5}$
Let,
Base =BC = 4k
Hypotenuse =AC = 5k
Where, k ia any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (4k)² = (5k)²
⇒ (AB)² + 16k² = 25k²
⇒ (AB)² = 25 k² –16 k²
⇒ (AB)² = 9 k²
⇒ AB =√9 k²
⇒ AB =±3k
But side AB can’t be negative. So, AB = 3k
Now, we have to find tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to angle θ = BC =4k
Side adjacent to angle θ = AB =3k
So,$\tan \theta=\frac{4 k}{3 k}=\frac{4}{3}$

Question 10 A 

If $\tan \theta=\frac{3}{4}$ then find the values of cos θ and sin θ.Sol :We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{3}{4}$$\Rightarrow \frac{P}{B}=\frac{3}{4}$$\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$
Let,
The side opposite to angle θ =AB = 3k
The side adjacent to angle θ =BC = 4k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9 k²+16 k²
⇒ (AC)² = 25 k²
⇒ AC =√25 k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AB = 3k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = 4k
and Hypotenuse = AC = 5k
So,$\cos \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

 Question 10 B 

If tan A= 4/3. Find the other trigonometric ratios of the angle A.Sol :

We know that,

$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Here, θ = A$\tan \mathrm{A}=\frac{4}{3}$$\Rightarrow \frac{P}{B}=\frac{4}{3}$$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}$Let,
The side opposite to angle A =BC = 4k
The side adjacent to angle A =AB = 3k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9 k² +16 k²
⇒ (AC)² = 25 k²
⇒ AC =√25 k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k
Now, we will find the sin A and cos A
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = 4k
and Hypotenuse = AC = 5k
So, $\sin A=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A = AB = 3k
and Hypotenuse = AC = 5k

So, $\cos A=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
Now, we find other trigonometric ratios
$\operatorname{cosec} \mathrm{A}=\frac{1}{\sin \mathrm{A}}$
$=\frac{1}{\frac{4}{5}}$
$=\frac{5}{4}$$\sec \mathrm{A}=\frac{1}{\cos \mathrm{A}}$
$=\frac{1}{\frac{3}{5}}$
$=\frac{5}{3}$$\cot A=\frac{1}{\tan A}$
$=\frac{1}{\frac{4}{3}}$
$=\frac{3}{4}$

Question 11 

If cot $\theta=\frac{12}{5}$, then find the value of sin θ.Sol :
We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{12}{5} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{12}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12}{5}$
Let,
Side adjacent to angle θ =AB = 12k
The side opposite to angle θ =BC = 5k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (5k)² = (AC)²
⇒ (AC)² = 144 k² +25 k²
⇒ (AC)² = 169 k²
⇒ AC =√169 k²
⇒ AC =±13k
But side AC can’t be negative. So, AC = 13k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 5k
and Hypotenuse = AC = 13k
So, $\sin \theta=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$

Question 12 

If tan $\theta=\frac{5}{12}$, then find the value of cos θ.Sol :
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{5}{12} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{5}{12} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$
Let,
The side opposite to angle θ =BC = 5k
The side adjacent to angle θ =AB = 12k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (5k)² = (AC)²
⇒ (AC)² = 144 k² +25 k²
⇒ (AC)² = 169 k²
⇒ AC =√169 k²
⇒ AC =±13k
But side AC can’t be negative. So, AC = 13k
Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB = 12k
and Hypotenuse = AC = 13k
So, $\cos \theta=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$

Question 13 

If sin $\theta=\frac{12}{13}$, then find the value of cos θ and tan θ.Sol :
Given: $\sin \theta=\frac{12}{13}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{12}{13} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{12}{13} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12}{13}$
Let,
Side opposite to angle θ = 12k
and Hypotenuse = 13k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (BCk)² = (13)²
⇒ 144 k² + (BC)² = 169 k²
⇒ (BC)² = 169 k² –144 k²
⇒ (BC)² = 25 k²
⇒ BC =√25 k²
⇒ BC =±5k
But side BC can’t be negative. So, BC = 5k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC =5k
Hypotenuse = AC =13k
So, $\cos \theta=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$
Now, tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
side opposite to angle θ = AB =12k
Side adjacent to angle θ = BC =5k
So, $\tan \theta=\frac{12 \mathrm{k}}{5 \mathrm{k}}=\frac{12}{5}$

Question 14 

If tan θ =0.75, then find the value of sin θ.Sol :We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$Given: tan θ =0.75
$\Rightarrow \tan \theta=\frac{75}{100}=\frac{3}{4}$
$\tan \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{3}{4} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}$
Let,
The side opposite to angle θ =BC = 3k
The side adjacent to angle θ =AB = 4k
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (3k)² = (AC)²
⇒ (AC)² = 16 k² +9 k²
⇒ (AC)² = 25 k²
⇒ AC =√25 k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 3k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{B C}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$

Question 15 

If tan B= √3, then find the values of sin B and cos B.Sol :We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: tan B = √3
$\Rightarrow \tan \mathrm{B}=\frac{\sqrt{3}}{1}$
$\tan \mathrm{B}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
Side opposite to angle B =AC = √3k
The side adjacent to angle B =AB = 1k
where k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k
Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle B = AC = k√3
and Hypotenuse = BC = 2k
So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle B = AB =1k
Hypotenuse = BC =2k
So, $\cos \mathrm{B}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$

Question 16 

If $\tan \theta=\frac{\mathrm{m}}{\mathrm{n}}$, then find the values of cos θ and sin θ.Sol :We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Here, $\tan \theta=\frac{\mathrm{m}}{\mathrm{n}}$
So, Side opposite to angle θ =AC = m
The side adjacent to angle θ =AB = n
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (n)² + (m)² = (BC)²
⇒ (BC)² = m² + n²
⇒ BC =√ m² + n²
So, BC =√(m² + n²)
Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = m
and Hypotenuse = BC =√(m² + n²)
So, $\sin \theta=\frac{A C}{B C}=\frac{m}{\sqrt{m^{2}+n^{2}}}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =n
Hypotenuse = BC =√(m² + n²)
So, $\cos \theta=\frac{A B}{B C}=\frac{n}{\sqrt{m^{2}+n^{2}}}$

Question 17 

If sin θ = √3 cos θ, then find the values of cos θ and sin θ.Sol :
Given : sin θ =√3cos θ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\sqrt{3}$
⇒ tan θ =√3
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
and tan θ = √3
$\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}$
$\tan \theta=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
The side opposite to angle θ =AC = k√3
The side adjacent to angle θ =AB = 1k
where k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (k√3)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = k√3
and Hypotenuse = BC = 2k
So,$\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =1k
Hypotenuse = BC =2k
So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$

Question 18 A 

If $\cot \theta=\frac{21}{20}$, then find the values of cos θ and sin θ.Sol :
We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{21}{20} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{21}{20} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{21}{20}$
Let,
Side adjacent to angle θ =AB = 21k
The side opposite to angle θ =BC = 20k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (21k)² + (20k)² = (AC)²
⇒ (AC)² = 441 k² +400 k²
⇒ (AC)² = 841 k²
⇒ AC =√841 k²
⇒ AC =±29k
But side AC can’t be negative. So, AC = 29k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 20k
and Hypotenuse = AC = 29k
So, $\sin \theta=\frac{B C}{A C}=\frac{20 k}{29 k}=\frac{20}{29}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =21k
Hypotenuse = AC =29k
So, $\cos \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{21 \mathrm{k}}{29 \mathrm{k}}=\frac{21}{29}$

Question 18 B 

If 15 cot A=18, find sin A and sec A.Sol :
Given: 15 cot A = 8
$\Rightarrow \cot A=\frac{8}{15}$
And we know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \mathrm{A}=\frac{8}{15} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{8}{15} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{8}{15}$
Let,
Side adjacent to angle A =AB = 8k
The side opposite to angle A =BC = 15k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (8k)² + (15k)² = (AC)²
⇒ (AC)² = 64 k² +225 k²
⇒ (AC)² = 289 k²
⇒ AC =√289 k²
⇒ AC =±17k
But side AC can’t be negative. So, AC = 17k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 15k
and Hypotenuse = AC = 17k
So, $\sin \theta=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =8
Hypotenuse = AC =17
So, $\cos \theta=\frac{A B}{A C}=\frac{8 k}{17 k}=\frac{8}{17}$
$\therefore \sec \theta=\frac{1}{\cos \theta}$$=\frac{1}{\frac{8}{17}}$$=\frac{17}{8}$

Question 19 

If sin θ = cos θ and 0° < θ <90°, then find the values of sin θ and cos θ.Sol :
Given: sinθ = cosθ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=1$
⇒ tan θ = 1$\tan \theta=\frac{1}{1} \Rightarrow \frac{P}{B}=\frac{1}{1} \Rightarrow \frac{A B}{B C}=\frac{1}{1}$
Let,
Side opposite to angle θ = AB =1k
The side adjacent to angle θ = BC =1k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (1k)² + (1k)² = (AC)²
⇒ (AC)² = 1k² +1k²
⇒ (AC)² = 2k²
⇒ AC =√2k²
⇒ AC =k√2
So, AC = k√2
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AB= 1k
and Hypotenuse = AC = k√2
So, $\sin \theta=\frac{A B}{A C}=\frac{1 k}{k \sqrt{2}}=\frac{1}{\sqrt{2}}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = BC =1k
Hypotenuse = AC =k√2
So, $\cos \theta=\frac{B C}{A C}=\frac{1 k}{k \sqrt{2}}=\frac{1}{\sqrt{2}}$

Question 20 

If $\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$, then find the values of cos θ and $\frac{1}{\tan \theta}$.Sol :
$\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+v^{2}}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or, $\sin \theta=\frac{\text { Perpendicular }}{\text { hypotenuse }}$$\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ $\Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{\mathrm{x}^{2}+\mathrm{y}^{2}}$ $ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{\mathrm{x}^{2}+\mathrm{y}^{2}}$Let,
Side opposite to angle θ = AB = x² – y²
and Hypotenuse = AC = x² + y²
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (x² – y² )² + (BC)² = (x² + y² )²
⇒ (BC)² = (x² + y² )² – (x² – y² )²
Using the identity, a² – b² = (a+b)(a – b)
⇒ (BC)² = [(x² + y² + x² – y² )][ x² + y² –( x² – y²)]
⇒ (BC)² = (2x²)(2y²)
⇒ (BC)² = (4x²y²)
⇒ BC =√4x²y²
⇒ BC = ±2xy
⇒ BC = 2xy [taking positive square root since, side cannot be negative]
$\therefore \cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{\text { BC }}{\text { AC }}=\frac{2 x y}{x^{2}+y^{2}}$
and $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}}$
So, $\frac{1}{\tan \theta}=\frac{1}{\frac{x^{2}-y^{2}}{2 x y}}=\frac{2 x y}{x^{2}-y^{2}}$

Question 21 

Iftan $\theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$, then find the values of sin θ and cos θ.Sol :
Given: $\tan \theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$
We know that,
$\tan \theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$
Let,
AB = √(m² – n²) and BC = n
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (√(m² – n²))² + (n)² = (AC )²
⇒ m² – n² + n² = (AC )²
⇒ (AC)² = (m²)
⇒ AC =√ m²
⇒ AC = ±m
⇒ AC = m [taking positive square root since, side cannot be negative]
Now, we have to find the value of cos θ and sin θ
We, know that
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{n}}{\mathrm{m}}$
and
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{\sqrt{m^{2}-n^{2}}}{m}$

Question 22 A 

If sec θ = 2, then find the values of other t–ratios of angle θ.Sol :
Given: sec θ = 2We know that,
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\operatorname{Sec} \theta=\frac{2}{1} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{2}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{2}{1}$
Let,
BC = 1k and AC = 2k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (1k)² = (2k )²
⇒ (AB)² + k² = 4k²
⇒ (AB)² = 4k² – k²
⇒ (AB)² = 3k²
⇒ AB = k√3
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{1 \mathrm{k}}=\frac{\sqrt{3}}{1}=\sqrt{3}$
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\sqrt{3}}$

Question 22 B 

Given $\sec \theta=\frac{13}{12}$ calculate all other trigonometric ratios.Sol :

Given: $\sec \theta=\frac{13}{12}$

We know that,
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\sec \theta=\frac{13}{12} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{13}{12} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13}{12}$
Let,
BC = 12k and AC = 13k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (12k)² = (13k )²
⇒ (AB)² + 144k² = 169k²
⇒ (AB)² = 169k² – 144k²
⇒ (AB)² = 25k²
⇒ AB = √25k²
⇒ AB =±5k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{B C}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{A B}{B C}=\frac{5 k}{12 k}=\frac{5}{12}$$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{5}{13}}=\frac{13}{5}$$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{5}{12}}=\frac{12}{5}$

Question 23 

If $\operatorname{cosec} \theta=\sqrt{10}$, then find the values of other t–ratios of angle θ.Sol :
We know that,
$\operatorname{cosec} \theta=\frac{\text { hypotenuse }}{\text { perpendicular }}$
$\operatorname{cosec} \theta=\frac{\sqrt{10}}{1} \Rightarrow \frac{\mathrm{H}}{\mathrm{P}}=\frac{\sqrt{10}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{10}}{1}$
Let,
AB = 1k and AC = k√10
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (1k )²+ (BC)² = (k√10)²
⇒ (BC)² = 10k² – k²
⇒ (BC)² = 9k²
⇒ BC = √9k²
⇒ BC =±3k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{\mathrm{k} \sqrt{10}}=\frac{1}{\sqrt{10}}$$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{B C}{A C}=\frac{3 k}{k \sqrt{10}}=\frac{3}{\sqrt{10}}$$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1 \mathrm{k}}{3 \mathrm{k}}=\frac{1}{3}$$\sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{\sqrt{10}}}=\frac{\sqrt{10}}{3}$$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{1}{3}}=3$

Question 24 A 

If $\tan \mathrm{A}=\frac{\sqrt{3}}{2}$ then find the values of sin A + cos A.Sol :We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: $\tan \mathrm{A}=\frac{\sqrt{3}}{2}$
$\Rightarrow \tan \mathrm{A}=\frac{\sqrt{3}}{2}$
$\tan \mathrm{A}=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\sqrt{3}}{2}$
Let,
Side opposite to angle A =BC = k√3
Side adjacent to angle A =AB = 2k
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (2k)² + (√3k)² = (AC)²
⇒ (AC)² = 4 k² +3 k²
⇒ (AC)² = 7 k²
⇒ AC =√7 k²
⇒ AC =k√7
So, AC = k√7

Now, we will find the sin A and cos A
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = k√3
and Hypotenuse = AC = k√7
So, $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{k} \sqrt{3}}{\mathrm{k} \sqrt{7}}=\frac{\sqrt{3}}{\sqrt{7}}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A = AB =2k
Hypotenuse = AC = k√7
So, $\cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{2 \mathrm{k}}{\mathrm{k} \sqrt{7}}=\frac{2}{\sqrt{7}}$
Now, we have to find sin A +cos A

Putting values of sin A and cos A, we get
$\sin \mathrm{A}+\cos \mathrm{A}=\frac{\sqrt{3}}{\sqrt{7}}+\frac{2}{\sqrt{7}}=\frac{\sqrt{3}+2}{\sqrt{7}}$

Question 24 B 

If $\sin \theta=\sqrt{3} \cos \theta$ find the value of cos θ – sin θ.Sol :
Given: sin θ =√3cos θ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\sqrt{3}$
⇒ tan θ = √3We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: tan θ = √3
$\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}$
$\tan \theta=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
Side opposite to angle θ =AC = √3k
Side adjacent to angle θ =AB = 1k
where k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k

Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = k√3
and Hypotenuse = BC = 2k
So, $\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =1k
Hypotenuse = BC =2k
So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$
Now, we have to find the value of cos θ – sin θ
Putting the values of sin θ and cos θ, we get
$\cos \theta-\sin \theta=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}$

Question 24 C 

If $\tan \theta=\frac{8}{15}$, find the value of 1+ cos² θ.Sol :We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{8}{15} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{8}{15} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{8}{15}$
Let,
Side opposite to angle θ =AB = 8k
Side adjacent to angle θ =BC = 15k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (8k)² + (15k)² = (AC)²
⇒ (AC)² = 64k²+225k²
⇒ (AC)² = 289 k²
⇒ AC =√289 k²
⇒ AC =±17k
But side AC can’t be negative. So, AC = 17k
Now, we will find the cos θ

We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = 15k
and Hypotenuse = AC = 17k
So, $\cos \theta=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}$
Now, we have to find the value of 1+ cos² θ

Putting the value of cos θ, we get
$1+\cos ^{2} \theta=1+\left(\frac{15}{17}\right)^{2}$$=1+\frac{225}{289}$$=\frac{289+225}{289}$$=\frac{514}{289}$

Question 25 

If $\cot \theta=\frac{7}{8}$, evaluate(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
(ii) $\cot ^{2} \theta$Sol :Given: $\cot \theta=\frac{7}{8}$
We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{7}{8} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{7}{8} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{7}{8}$
Let,
Side adjacent to angle θ =AB = 7k
Side opposite to angle θ =BC = 8k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (7k)² + (8k)² = (AC)²
⇒ (AC)² = 49 k² +69 k²
⇒ (AC)² = 113 k²
⇒ AC =√113 k²
⇒ AC =k√113
$\therefore \sin \theta=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{8 \mathrm{k}}{\mathrm{k} \sqrt{113}}=\frac{8}{\sqrt{113}}$and $\cos \theta=\frac{B}{H}=\frac{A B}{A C}=\frac{7 k}{k \sqrt{113}}=\frac{7}{\sqrt{113}}$(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
We know that,
(a+b)(a – b) = (a² – b²)
So, using this identity, we get
$=\frac{(1)^{2}-(\sin \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$$=\frac{1-\sin ^{2} \theta}{1-\cos ^{2} \theta}$$=\frac{1-\left(\frac{8}{\sqrt{113}}\right)^{2}}{1-\left(\frac{7}{\sqrt{113}}\right)^{2}}$$=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}$$=\frac{\frac{113-64}{113}}{\frac{113-49}{113}}$$=\frac{49}{64}$(ii) cot² θ
Given $\cot \theta=\frac{7}{8}$$=\left(\frac{7}{8}\right)^{2}$$=\frac{49}{64}$

Question 26 A 

If 3 cot A = 4, check whether $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$=cos²A–sin² A or not.Sol :
Given: 3cot A = 4
$\Rightarrow \cot A=\frac{4}{3}$We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot A=\frac{4}{3} \Rightarrow \frac{B}{P}=\frac{4}{3} \Rightarrow \frac{A B}{B C}=\frac{4}{3}$
Let,
Side adjacent to angle A =AB = 4k
The side opposite to angle A =BC = 3k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (3k)² = (AC)²
⇒ (AC)² = 16 k² + 9 k²
⇒ (AC)² = 25 k²
⇒ AC =√25k²
⇒ AC = ±5k [taking positive square root since, side cannot be negative]
$\therefore \tan \mathrm{A}=\frac{1}{\cot \mathrm{A}}=\frac{1}{\frac{4}{3}}=\frac{3}{4}$
$\sin \mathrm{A}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$
and $\cos A=\frac{B}{H}=\frac{A B}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, $\mathrm{LHS}=\frac{1-\tan ^{2} \mathrm{A}}{1+\tan ^{2} \mathrm{A}}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}$$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$$=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}$$=\frac{7}{25}$ …(i)

And RHS = cos² A – sin² A
$=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}$$=\frac{16}{25}-\frac{9}{25}$$=\frac{7}{25}$ …(ii)

From Eqs. (i) and (ii) LHS =RHS
Hence Proved

Question 26 B 

In a right triangle ABC, right angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.Sol :
tan A = 1
As we know
$\tan \theta=\frac{\text { perpedicular }}{\text { base }}$
Now construct a right angle triangle right angled at B such that
∠ BAC = θ
Hence perpendicular = BC = 1 and base = AB = 1By Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = (1)² + (1)²
⇒ AC² = 2
⇒ AC = 
As,$\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}$
 and $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
⇒ $\sin \theta=\frac{1}{\sqrt{2}}$ and $\cos \theta=\frac{1}{\sqrt{2}}$
Hence,
2 sin A cos A=$2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
⇒ 2 sin A cos A=$2 \times \frac{1}{2}$
⇒ 2 sin A cos A=1
= R.H.S
Hence proved.

Question 27 

If 4sin² θ =3 and 0^o < θ <90^o, find the value of 1 + cos θ.Sol :
4sin² θ =3
$\Rightarrow \sin ^{2} \theta=\frac{3}{4}$
$\Rightarrow \sin \theta=\pm \frac{\sqrt{3}}{2}$

But it is given 0^o< θ <90^o
So, $\sin \theta=\frac{\sqrt{3}}{2}$
$\sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\sqrt{3}}{2}$

Let, P =k√3 and H =2k
In right angled ∆ABC, we have
B² + P² = H²
⇒ B² + (k√3)² = (2k)²
⇒ B² + 3k² = 4k²
⇒ B² = 4k² – 3k²
⇒ B² = k²
⇒ B = ±k
⇒ B = k [taking positive square root since, side cannot be negative]
$\therefore \cos \theta=\frac{B}{H}=\frac{k}{2 k}=\frac{1}{2}$
So, $1+\cos \theta=1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}$

Question 28 

If $\tan \theta=\frac{p}{q}$find the value of $\frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}$.Sol :
Given:$\tan \theta=\frac{p}{q}$

Now, $\frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}$
$\Rightarrow \frac{\cos \theta\left(\frac{p \sin \theta}{\cos \theta}-q\right)}{\cos \theta\left(\frac{p \sin \theta}{\cos \theta}+q\right)}$$\Rightarrow \frac{p \tan \theta-q}{p \tan \theta+q}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$$\Rightarrow \frac{p\left(\frac{p}{q}\right)-q}{p\left(\frac{p}{q}\right)+q}$
$\Rightarrow \frac{\frac{p^{2}-q^{2}}{q}}{\frac{p^{2}+q^{2}}{q}}$
$\Rightarrow \frac{p^{2}-q^{2}}{p^{2}+q^{2}}$
$\therefore \frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}=\frac{p^{2}-q^{2}}{p^{2}+q^{2}}$

Question 29 

If 13 cos θ = 5, $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$.Sol :
Given: 13 cosθ = 5
$\Rightarrow \cos \theta=\frac{5}{13}$We know that,
$\cos \theta=\frac{\text { Base }}{\text { hypotenuse }}$
$\cos \theta=\frac{5}{13} \Rightarrow \frac{B}{H}=\frac{5}{13}$Let AB =5k and BC = 13k
In right angled ∆ABC, we have
B² + P² = H²
⇒ (5k)² + P² = (13k)²
⇒ P² + 25k² = 169k²
⇒ P² = 169k² – 25k²
⇒ P² = 144k²
⇒ P =√144k²
⇒ P = ±12k
⇒ P = 12k [taking positive square root since, side cannot be negative]
$\therefore \sin \theta=\frac{P}{H}=\frac{12}{13}$Now, $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$
$=\frac{\frac{12}{13}+\frac{5}{13}}{\frac{12}{13}-\frac{5}{13}}$
$=\frac{17}{7}$

Question 30 

If $\sec \theta=\frac{13}{5}$, show that $\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}=3$.Sol :
Given: $\sec \theta=\frac{13}{5}$We know that,
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\operatorname{Sec} \theta=\frac{13}{5} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{13}{5} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13}{5}$
Let,
BC = 5k and AC = 13k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (5k)² = (13k )²
⇒ (AB)² + 25k² = 169k²
⇒ (AB)² = 169k² – 25k²
⇒ (AB)² = 144k²
⇒ AB = √144k²
⇒ AB =±12k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$

Now, LHS $=\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}$
$=\frac{2\left(\frac{12}{13}\right)-3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right)-9\left(\frac{5}{13}\right)}$$=\frac{24-15}{48-45}$$=\frac{9}{3}$=3 =RHS
Hence Proved

Question 31 

If 2 tan θ = 1, find the value of $\frac{3 \cos \theta+\sin \theta}{2 \cos \theta-\sin \theta}$.Sol :
Given: 2 tan θ = 1
$\Rightarrow \tan \theta=\frac{1}{2}$We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{1}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{1}{2} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{2}$
Let,
Side opposite to angle θ =AB = 1k
Side adjacent to angle θ =BC = 2k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (k)² + (2k)² = (AC)²
⇒ (AC)² = k²+4k²
⇒ (AC)² = 5k²
⇒ AC =√5k²
⇒ AC =±k√5
But side AC can’t be negative. So, AC = k√5

Now, we will find the sin θ and cos θ

We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = 2k
and Hypotenuse = AC = k√5
So,$\cos \theta=\frac{B C}{A C}=\frac{2 k}{k \sqrt{5}}=\frac{2}{\sqrt{5}}$
And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ =AB = 1k
And Hypotenuse =AC = k√5
So, $\sin \theta=\frac{A B}{A C}=\frac{1 k}{k \sqrt{5}}=\frac{1}{\sqrt{5}}$
Now, $\frac{3 \cos \theta+\sin \theta}{2 \cos \theta-\sin \theta}$
$=\frac{3\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}{2\left(\frac{2}{\sqrt{5}}\right)-\frac{1}{\sqrt{5}}}$

$=\frac{6+1}{4-1}$

$=\frac{7}{3}$

Question 32 

If 5 tan α = 4, show that $\frac{5 \sin \alpha-3 \cos \alpha}{5 \sin \alpha+2 \cos \alpha}=\frac{1}{6}$.Sol :
Given: 5 tan = 4
$\Rightarrow \tan \alpha=\frac{4}{5}$We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \alpha=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{4}{5}$Let,
The side opposite to angle α =AB = 4k
The side adjacent to angle α =BC = 5k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (5k)² = (AC)²
⇒ (AC)² = 16k²+25k²
⇒ (AC)² = 41k²
⇒ AC =√41k²
⇒ AC =±k√41
But side AC can’t be negative. So, AC = k√41
Now, we will find the sin α and cos α
We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle α = BC = 5k
and Hypotenuse = AC = k√41
So, $\cos \alpha=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 \mathrm{k}}{\mathrm{k} \sqrt{41}}=\frac{5}{\sqrt{41}}$
And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle α =AB = 4k
And Hypotenuse =AC = k√5
So, $\sin \alpha=\frac{A B}{A C}=\frac{4 k}{k \sqrt{4} 1}=\frac{4}{\sqrt{41}}$

Now, LHS $=\frac{5 \sin \alpha-3 \cos \alpha}{5 \sin \alpha+2 \cos \alpha}$
$=\frac{5\left(\frac{4}{\sqrt{41}}\right)-3\left(\frac{5}{\sqrt{41}}\right)}{5\left(\frac{4}{\sqrt{41}}\right)+2\left(\frac{5}{\sqrt{41}}\right)}$
$=\frac{20-15}{20+10}$$=\frac{5}{30}$$=\frac{1}{6}$ = RHS
Hence Proved

Question 33 

If $\cot \theta=\frac{3}{4}$prove that $\sqrt{\frac{\sec \theta+\operatorname{cosec} \theta}{\sec \theta-\operatorname{cosec} \theta}}=\sqrt{7}$.Sol :We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{3}{4} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$
Let,
Side adjacent to angle θ =AB = 3k
The side opposite to angle θ =BC = 4k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9k² +16k²
⇒ (AC)² = 25k²
⇒ AC =√25k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 4k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =3k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
$\therefore \sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$
And
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{4}{5}}=\frac{5}{4}$

Now, LHS$=\sqrt{\frac{\sec \theta+\operatorname{cosec} \theta}{\sec \theta-\operatorname{cosec} \theta}}$$=\sqrt{\frac{\left(\frac{5}{3}\right)+\left(\frac{5}{4}\right)}{\left(\frac{5}{3}\right)-\left(\frac{5}{4}\right)}}$$=\sqrt{\frac{\left(\frac{20+15}{12}\right)}{\left(\frac{20-15}{12}\right)}}$$=\sqrt{\frac{35}{5}}$$=\sqrt{7}$ = RHS
Hence Proved

Question 34

If $\cot \theta=\frac{1}{\sqrt{3}}$ verify that: $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{3}{5}$.Sol :We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1}{\sqrt{3}}$Let,
Side adjacent to angle θ =AB = 1k
Side opposite to angle θ =AC = k√3
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = k√3
and Hypotenuse = BC = 2k
So$\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =1k
Hypotenuse = BC =2k
So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$

Now, LHS $=\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$
$=\frac{1-\left(\frac{1}{2}\right)^{2}}{2-\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}$
$=\frac{\frac{4-1}{4}}{\frac{8-3}{4}}$
$=\frac{3}{5}$ = RHS
Hence Proved

Question 35 

If $\tan \theta=\frac{x}{y}$ find the value of x sin θ + y cos θ.Sol :

We know that,

$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{x}{y} \Rightarrow \frac{P}{B}=\frac{x}{y} \Rightarrow \frac{A B}{B C}=\frac{x}{y}$
Let,
Side opposite to angle θ =AB = x
Side adjacent to angle θ =BC = y
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (x)² + (y)² = (AC)²
⇒ (AC)² = x²+y²
⇒ AC =√( x²+y²)
Now, we will find the sin θ and cos θ
We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = y
and Hypotenuse = AC = √( x²+y²)
So,$\cos \theta=\frac{B C}{A C}=\frac{y}{\sqrt{x^{2}+y^{2}}}$
And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ =AB = x
And Hypotenuse =AC = √( x²+y²)
So, $\sin \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}$
Now, x sin θ +y cos θ
$=\mathrm{x}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}\right)+\mathrm{y}\left(\frac{\mathrm{y}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}\right)$
$=\frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}}}$
= √( x²+y²)

Question 36 

If $\sin \theta=\frac{3}{5}$, find the value of tan²θ + sinθ cosθ + cotθ.Given: $\sin \theta=\frac{3}{5}$We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypollyuse }}$
$\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$
Let,
Perpendicular =AB =3k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (BC)² = (5k)²
⇒ 9k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –9k²
⇒ (BC)² = 16k²
⇒ BC =√16k²
⇒ BC =±4k
But side BC can’t be negative. So, BC = 4k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
The side adjacent to angle θ or base = BC =4k
Hypotenuse = AC =5k
So,$\cos \theta=\frac{4 k}{5 k}=\frac{4}{5}$
Now,
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =3k
Base = BC =4k
So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
Now, tan² θ + sin θ cos θ + cot θ
$=\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)+\left(\frac{4}{3}\right)$$=\left(\frac{9}{16}\right)+\left(\frac{13}{25}\right)+\left(\frac{4}{3}\right)$$=\frac{675+576+1600}{16 \times 25 \times 3}$$=\frac{2851}{1200}$

Question 37 

If 4cot θ = 3, show that $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=7$.Sol :
Given: $\cot \theta=\frac{3}{4}$We know that,$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{3}{4} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$
Let,
Side adjacent to angle θ =AB = 3k
The side opposite to angle θ =BC = 4k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9k² +16k²
⇒ (AC)² = 25k²
⇒ AC =√25k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 4k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =3k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$

Now, LHS $=\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$
$=\frac{\frac{4}{5}+\frac{3}{5}}{\frac{4}{5}-\frac{3}{5}}$$=\frac{7}{1}$= 7 = RHS
Hence Proved

Question 38 

If $\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$, prove that $\mathrm{m} \sin \theta+\mathrm{n} \cos \theta=\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}$Sol :
Given: $\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$

Let,
Perpendicular =AB =m
and Hypotenuse =AC =√(m² + n²)
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (m)² + (BC)² = (√(m² + n²))²
⇒ m² + (BC)² = m² + n²
⇒ (BC)² = m² + n² – m²
⇒ (BC)² = n²
⇒ BC =√n²
⇒ BC =±n
But side BC can’t be negative. So, BC = n
Now, we have to find the value of cos θ and tan θ

We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
Side adjacent to angle θ or base = BC =n
Hypotenuse = AC =√(m² + n²)
So, $\cos \theta=\frac{n}{\sqrt{m^{2}+n^{2}}}$

Now, LHS = m sin θ +n cosθ
$=\mathrm{m}\left(\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}\right)+\mathrm{n}\left(\frac{\mathrm{n}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}\right)$
$=\frac{\mathrm{m}^{2}+\mathrm{n}^{2}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$
=√(m² + n²) = RHS
Hence Proved

Question 39 

If $\cos \alpha=\frac{12}{13}$ show that $\sin \alpha(1-\tan \alpha)=\frac{35}{156}$Sol :
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$
$\cos \alpha=\frac{12}{13} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{12}{13} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{12}{13}$

Let,
Base =BC = 12k
Hypotenuse =AC = 13k
Where, k ia any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (12k)² = (13k)²
⇒ (AB)² + 144k² = 169k²
⇒ (AB)² = 169 k² –144 k²
⇒ (AB)² = 25 k²
⇒ AB =√25 k²
⇒ AB =±5k
But side AB can’t be negative. So, AB = 5k
Now, we have to find sin α and tan α

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle α = AB =5k
And Hypotenuse = AC =13k
So, $\sin \alpha=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to angle α = AB =5k
Side adjacent to angle α = BC =12k
So, $\tan \alpha=\frac{5 \mathrm{k}}{12 \mathrm{k}}=\frac{5}{12}$
Now, LHS = sin α (1 – tan α)

$=\frac{5}{13}\left(1-\frac{5}{12}\right)$

$=\frac{5}{13}\left(\frac{12-5}{12}\right)$

$=\frac{35}{156}$ = RHS
Hence Proved

Question 40 

If $\mathrm{q} \cos \theta=\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}$, prove that q sin θ = p.Sol :
Given : q cos θ = √(q² – p²)
$\Rightarrow \cos \theta=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}}$

We know that,

$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$
$\cos \theta=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}}$

Let,
Base =BC = √(q² – p²)
Hypotenuse =AC = q
Where, k ia any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (√(q² – p²))² = (q)²
⇒ (AB)² + (q² – p²) = q²
⇒ (AB)² = q² – q² + p²)
⇒ (AB)² = p²
⇒ AB =√p²
⇒ AB =±p
But side AB can’t be negative. So, AB = p
Now, we have to find sin θ

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
The side opposite to angle θ = AB =p
And Hypotenuse = AC =q
So, $\sin \theta=\left(\frac{p}{q}\right)$
Now, LHS = q sin θ
$=\mathrm{q}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)$
= q = RHS
Hence Proved

Question 41 

If $\sin \theta=\frac{3}{5}$, show that : $\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}=-\frac{1}{5}$Sol :
Given: $\sin \theta=\frac{3}{5}$

We know that,

$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$
Let,
Perpendicular =AB =3k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (BC)² = (5k)²
⇒ 9k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –9k²
⇒ (BC)² = 16k²
⇒ BC =√16k²
⇒ BC =±4k
But side BC can’t be negative. So, BC = 4k
Now, we have to find the value of cos θ and tan θ

We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
The side adjacent to angle θ or base = BC =4k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{4 k}{5 k}=\frac{4}{5}$
Now,
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =3k
Base = BC =4k
So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
Now, LHS $=\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}$
$=\frac{\left(\frac{4}{5}\right)-\left(\frac{1}{\frac{3}{4}}\right)}{2\left(\frac{4}{3}\right)}$
$=\frac{\left(\frac{4}{5}\right)-\left(\frac{4}{3}\right)}{\left(\frac{8}{3}\right)}$
$=\frac{\frac{12-20}{15}}{\left(\frac{8}{3}\right)}$
$=\frac{\left(-\frac{8}{15}\right)}{\left(\frac{8}{3}\right)}$
$=-\frac{1}{5}$ = RHS
Hence Proved

Question 42 A 

Find the value ofcos A sin B + sin A. cos B, if sin A= 4/5 and cos B = 12/13.
Sol :

Given: $\sin A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$
To find: cos A sin B + sin A cos B
As, we have the value of sin A and cos B but we don’t have the value of cos A and sin B
So, First we find the value of cos A and sin B
$\sin A=\frac{4}{5}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$\sin A=\frac{4}{5} \Rightarrow \frac{P}{H}=\frac{4}{5}$
Let,
Side opposite to angle A = 4k
and Hypotenuse = 5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (P)² + (B)² = (H)²
⇒ (4k)² + (B)² = (5)²
⇒ 16 k² + (B)² = 25 k²
⇒ (B)² = 25 k² –16 k²
⇒ (B)² = 9 k²
⇒ B =√9 k²
⇒ B =±3k [taking positive square root since, side cannot be negative]
So, Base = 3k
Now, we have to find the value of cos A
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A =3k
Hypotenuse =5k
So, $\cos \mathrm{A}=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$
Now, we have to find the sin B

We know that,

$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
$\cos \mathbf{B}=\frac{12}{13} \Rightarrow \frac{B}{H}=\frac{12}{13}$
Let,
Side adjacent to angle B =12k
Hypotenuse =13k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (B)² + (P)² = (H)²
⇒ (12k)² + (P)² = (13)²
⇒ 144 k² + (P)² = 169 k²
⇒ (P)² = 169 k² –144 k²
⇒ (P)² = 25 k²
⇒ P =√25 k²
⇒ P =±5k [taking positive square root since, side cannot be negative]
So, Perpendicular = 5k
Now, we have to find the value of sin B

We know that,
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{B}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$
Now, cos A sin B + sin A cos B

Putting the values of sin A, sin B cos A and Cos B, we get

$\Rightarrow\left(\frac{3}{5}\right)\left(\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)$

$\Rightarrow \frac{15+48}{5 \times 13}$

$\Rightarrow \frac{63}{65}$

Question 42 B 

Find the value ofsin A. cos B – cos A. sin B, if tan A= √3 and sin B = 1/2.
Sol :

Given: tan A =√3 and $\sin B=\frac{1}{2}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{B}=\frac{1}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{1}{2}$
Let,
Side opposite to angle θ = 1k
and Hypotenuse = 2k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AC)² + (BC)² = (AB)²
⇒ (1k)² + (BC)² = (2k)²
⇒ k² + (BC)² = 4k²
⇒ (BC)² = 4k² –k²
⇒ (BC)² = 3 k²
⇒ BC =√3k²
⇒ BC =k√3
So, BC = k√3

Now, we have to find the value of cos B
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle B = BC =k√3
Hypotenuse = AB =2k
So, $\cos \mathbf{B}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: tan A = √3
$\Rightarrow \tan \mathrm{A}=\frac{\sqrt{3}}{1}$
$\tan \mathrm{A}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
The side opposite to angle A =BC = √3k
The side adjacent to angle A =AB = 1k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (1k)² + (√3k)² = (AC)²
⇒ (AC)² = 1 k² +3 k²
⇒ (AC)² = 4 k²
⇒ AC =√2 k²
⇒ AC =±2k
But side AC can’t be negative. So, AC = 2k
Now, we will find the sin A and cos A
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = k√3
and Hypotenuse = AC = 2k
So, $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle A = AB =1k
Hypotenuse = AC =2k

So, $\cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$
Now, sin A. cos B – cos A. sin B

Putting the values of sin A, sin B cos A and Cos B, we get

$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$

$\Rightarrow \frac{3}{4}-\frac{1}{4}$

$\Rightarrow \frac{2}{4}$

$=\frac{1}{2}$

Question 42 C 

Find the value ofsin A. cos B + cos A. sin B. if $\tan \mathrm{A}=\frac{1}{\sqrt{3}}$and tan B = √3.
Sol :
Given:

$\tan \mathrm{A}=\frac{1}{\sqrt{3}}$
$\tan \mathrm{A}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{1}{\sqrt{3}}$
Let,
Side opposite to angle A =BC = 1k
Side adjacent to angle A =AB = k√3
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (√3k)² + (1k)² = (AC)²
⇒ (AC)² = 1 k² +3 k²
⇒ (AC)² = 4 k²
⇒ AC =√2 k²
⇒ AC =±2k
But side AC can’t be negative. So, AC = 2k

Now, we will find the sin A and cos A

$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = k
and Hypotenuse = AC = 2k
So, $\operatorname{Sin} \mathbf{A}=\frac{B C}{A C}=\frac{1 k}{2 k}=\frac{1}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A = AB =k√3
Hypotenuse = AC =2k
So, $\cos \mathbf{A}=\frac{A B}{B C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$
Now,

Given: tan B = √3
$\Rightarrow \tan \mathrm{B}=\frac{\sqrt{3}}{1}$
$\tan \mathrm{B}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
Side opposite to angle B =AC = √3k
Side adjacent to angle B =AB = 1k
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k

Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle B = AC = k√3
and Hypotenuse = BC = 2k
So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle B = AB =1k
Hypotenuse = BC =2k
So, $\cos \mathbf{B}=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$
Now, sin A. cos B + cos A. sin B

Putting the values of sin A, sin B cos A and Cos B, we get

$\Rightarrow\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$

$\Rightarrow \frac{1}{4}+\frac{3}{4}$

$\Rightarrow \frac{4}{4}$

=1

Question 42 D 

Find the value of$\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}$, if sin A = $\frac{1}{\sqrt{2}}$and cos $\mathrm{B}=\frac{\sqrt{3}}{2}$
Sol :
Given $: \sin \mathrm{A}=\frac{1}{\sqrt{2}}$ and $\cos \mathrm{B}=\frac{\sqrt{3}}{2}$
$\sin \mathrm{A}=\frac{1}{\sqrt{2}}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{A}=\frac{1}{\sqrt{2}} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{1}{\sqrt{2}}$

Let,
Side opposite to angle A = k
and Hypotenuse = k√2
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (P)² + (B)² = (H)²
⇒ (k)² + (B)² = (k√2)²
⇒ k² + (B)² = 2k²
⇒ (B)² = 2k² – k²
⇒ (B)² = k²
⇒ B =√k²
⇒ B =±k [taking positive square root since, side cannot be negative]
So, Base = k
Now, we have to find the value of tan A

We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
So, $\tan \mathrm{A}=\frac{\mathrm{k}}{\mathrm{k}}=1$

Now, we have to find the tan B
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
$\cos \mathbf{B}=\frac{\sqrt{3}}{2} \Rightarrow \frac{B}{H}=\frac{\sqrt{3}}{2}$
Let,
Side adjacent to angle B =k√3
Hypotenuse =2k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (B)² + (P)² = (H)²
⇒ (k√3)² + (P)² = (2k)²
⇒ 3k² + (P)² = 4k²
⇒ (P)² = 4k² –3 k²
⇒ (P)² = k²
⇒ P =√k²
⇒ P =±k [taking positive square root since, side cannot be negative]
So, Perpendicular = k
Now, we have to find the value of sin B
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
So, $\tan \mathrm{B}=\frac{\mathrm{k}}{\mathrm{k} \sqrt{3}}=\frac{1}{\sqrt{3}}$

Now, $\frac{\tan A+\tan B}{1-\tan A \tan B}$
$\Rightarrow \frac{(1)+\left(\frac{1}{\sqrt{3}}\right)}{1-(1)\left(\frac{1}{\sqrt{3}}\right)}$

$\Rightarrow \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}$

$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1}$

Now, multiply and divide by the conjugate of √3 – 1, we get
$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\Rightarrow \frac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$ [∵ (a – b)(a+b) = (a² – b²)]

$\Rightarrow \frac{3+1+2 \sqrt{3}}{3-1}$

$\Rightarrow \frac{4+2 \sqrt{3}}{2}$

⇒ 2+√3

Question 42 E

Find the value ofsec A. tan A+tan²A – cosec A, if tan A =2
Sol :
Given: tan A = 2 ⇒ tan²A = 4
We know that, sec² A = 1+ tan²A
⇒ sec² A = 1 + 4
⇒ sec² A = 5
⇒ sec A =√5
$\Rightarrow \cos A=\frac{1}{\sqrt{5}}$

Now, we know that tan A$=\frac{\sin A}{\cos A}$
$\Rightarrow 2=\frac{\sin A}{\frac{1}{\sqrt{5}}}$
⇒ 2 =√5 sin A

$\Rightarrow \sin A=\frac{2}{\sqrt{5}}$

$\Rightarrow \operatorname{cosec} A=\frac{\sqrt{5}}{2}$

Now, putting all the values in the given equation, we get
sec A. tan A+tan²A – cosec A

$\Rightarrow(\sqrt{5})(2)+(4)-\left(\frac{\sqrt{5}}{2}\right)$

$\Rightarrow \frac{4 \sqrt{5}+8-\sqrt{5}}{2}$

$\Rightarrow \frac{3 \sqrt{5}+8}{2}$

Question 42 F

Find the value of$\frac{1}{\tan \mathrm{A}}+\frac{\sin \mathrm{A}}{1+\cos \mathrm{A}}$, if cosec A = 2
Sol :
Given: cosec A =2
Now, we have to find $\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}$
First, we simplify the above given trigonometry equation, we get
$\frac{1}{\frac{\sin A}{\cos A}}+\frac{\sin A}{1+\cos A}$
$\Rightarrow \frac{\cos A}{\sin A}+\frac{\sin A}{1+\cos A}$

Taking the LCM, we get
$\Rightarrow \frac{\cos A(1+\cos A)+\sin A(\sin A)}{(\sin A)(1+\cos A)}$
$\Rightarrow \frac{\cos A+\cos ^{2} A+\sin ^{2} A}{\sin A(1+\cos A)}$ [∵ cos²θ +sin² θ = 1]
$\Rightarrow \frac{\cos A+1}{\sin A(1+\cos A)}$
$\Rightarrow \frac{1}{\sin A}$ [∵ cosec θ $=\frac{1}{\sin \theta}$ ]
⇒ cosec A
⇒ 2

Question 43 A

If $\sin \mathrm{B}=\frac{1}{2}$, prove that : 3 cos B – 4cos³ B = 0Sol :
Given: $\sin B=\frac{1}{2}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin B=\frac{1}{2} \Rightarrow \frac{P}{H}=\frac{1}{2} \Rightarrow \frac{A B}{A C}=\frac{1}{2}$
Let,
Perpendicular =AB =k
and Hypotenuse =AC =2k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (k)² + (BC)² = (2k)²
⇒ k² + (BC)² = 4k²
⇒ (BC)² = 4k² –k²
⇒ (BC)² = 3k²
⇒ BC =√3k²
⇒ BC =k√3
So, BC = k√3
Now, we have to find the value of cos B

We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
Side adjacent to angle B or base = BC = k√3
Hypotenuse = AC =2k
So, $\cos \mathrm{B}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, LHS = 3 cos B – 4cos³ B

$\Rightarrow 3\left(\frac{\sqrt{3}}{2}\right)-4\left(\frac{\sqrt{3}}{2}\right)^{3}$

$\Rightarrow \frac{3 \sqrt{3}}{2}-4\left(\frac{3 \sqrt{3}}{8}\right)$

$\Rightarrow \frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}=0$

=RHS
Hence Proved

Question 43 B

If $\cos \theta=\frac{\sqrt{3}}{2}$, prove that: 3sinθ – 4sin³θ = 1.Sol :

We know that,

$\cos \theta=\frac{\text { Base }}{\text { hypotenuse }}$
$\cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{\sqrt{3}}{2}$
Let AB =k√3 and BC = 2k
In right angled ∆ABC, we have
B² + P² = H²
⇒ (k√3)² + P² = (2k)²
⇒ P² + 3k² = 4k²
⇒ P² = 4k² – 3k²
⇒ P² = k²
⇒ P =√k²
⇒ P = ±k
⇒ P = k [taking positive square root since, side cannot be negative]
Now,
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\therefore \sin \theta=\frac{P}{H}=\frac{k}{2 k}=\frac{1}{2}$

Now, LHS = 3sin θ – 4sin³ θ

$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^{3}$

$\Rightarrow \frac{3}{2}-\frac{4}{8}$

$\Rightarrow \frac{3}{2}-\frac{1}{2}$

$\Rightarrow \frac{3-1}{2}$

$\Rightarrow \frac{2}{2}$

⇒ 1 = RHS
Hence Proved

Question 43 C

If $\sec \theta=\frac{5}{4}$, prove that : $\frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\sin \theta}{\sec \theta}$Sol :
Given: $\sec \theta=\frac{5}{4}$

We know that,

$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\operatorname{Sec} \theta=\frac{5}{4} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{5}{4} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{5}{4}$
Let,
BC = 4k and AC = 5k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (4k)² = (5k )²
⇒ (AB)² + 16k² = 25k²
⇒ (AB)² = 25k² – 16k²
⇒ (AB)² = 9k²
⇒ AB = √9k²
⇒ AB =±3k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that

$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

$=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$=\frac{A B}{B C}=\frac{3 k}{4 k}=\frac{3}{4}$

Now, LHS $=\frac{\tan \theta}{1+\tan ^{2} \theta}$

$=\frac{\left(\frac{3}{4}\right)}{1+\left(\frac{3}{4}\right)^{2}}$

$=\frac{\left(\frac{3}{4}\right)}{1+\left(\frac{9}{16}\right)}$

$=\frac{\left(\frac{3}{4}\right)}{\left(\frac{16+9}{16}\right)}$

$=\frac{\left(\frac{3}{4}\right)}{\left(\frac{25}{16}\right)}$

$=\frac{3}{4} \times \frac{16}{25}$

$=\frac{12}{25}$

Now, RHS $=\frac{\sin \theta}{\sec \theta}$

$=\frac{\left(\frac{3}{5}\right)}{\left(\frac{5}{4}\right)}$

$=\frac{3}{5} \times \frac{4}{5}$

$=\frac{12}{25}$

∴ LHS = RHS
Hence Proved

Question 43 D

$\cot B=\frac{12}{5}$, prove that : tan²B – sin² B=sin⁴ B sec² B.Sol :

We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot B=\frac{12}{5} \Rightarrow \frac{B}{P}=\frac{12}{5} \Rightarrow \frac{A B}{B C}=\frac{12}{5}$
Let,
Side adjacent to angle B =AB = 12k
Side opposite to angle B =BC = 5k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (5k)² = (AC)²
⇒ (AC)² = 144 k² +25 k²
⇒ (AC)² = 169 k²
⇒ AC =√169 k²
⇒ AC =±13k
But side AC can’t be negative. So, AC = 13k

Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle B = BC = 5k
and Hypotenuse = AC = 13k
So, $\sin \mathrm{B}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle B = AB =12k
Hypotenuse = AC =13k
So, $\cos \mathrm{B}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 \mathrm{k}}{13 \mathrm{k}}=\frac{12}{13}$
$\tan \mathrm{B}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$
$\sec \mathrm{B}=\frac{1}{\cos \mathrm{B}}=\frac{1}{\frac{12}{13}}=\frac{13}{12}$

Now, LHS = tan²B – sin² B

$=\left(\frac{5}{12}\right)^{2}-\left(\frac{5}{13}\right)^{2}$

$=\frac{25}{144}-\frac{25}{169}$

$=\frac{4225-3600}{144 \times 169}$

$=\frac{625}{144 \times 169}$

$=\frac{625}{24336}$

Now, RHS = sin⁴ B sec² B

$=\left(\frac{5}{13}\right)^{4}\left(\frac{13}{12}\right)^{2}$

$=\left(\frac{5}{13}\right)^{2}\left(\frac{5}{13}\right)^{2}\left(\frac{13}{12}\right)^{2}$

$=\frac{625}{144 \times 169}$

$=\frac{625}{24336}$

Now, LHS = RHS
Hence Proved

Question 44 

If $\cos \theta=\frac{\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}$, prove that $\left(\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}= \frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}+\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}-\mathrm{q}}}$Sol :
Given: $\cos \theta=\frac{q}{\sqrt{p^{2}+q^{2}}}$
Now, squaring both the sides, we get

$=\cos ^{2} \theta=\frac{\mathrm{q}^{2}}{\mathrm{p}^{2}+\mathrm{q}^{2}}$

$\Rightarrow \mathrm{p}^{2}+\mathrm{q}^{2}=\frac{\mathrm{q}^{2}}{\cos ^{2} \theta}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2}}{\cos ^{2} \theta}-\mathrm{q}^{2}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2}-\mathrm{q}^{2} \cos ^{2} \theta}{\cos ^{2} \theta}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2}\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2} \sin ^{2} \theta}{\cos ^{2} \theta}$

⇒ p² = q² tan²θ …(1)

Now, solving LHS $=\left(\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}$
Putting the value of p² in the above equation, we get

$=\left(\frac{\sqrt{q^{2} \tan ^{2} \theta+q^{2}}}{p}+\frac{q}{p}\right)^{2}$

$\Rightarrow\left(\frac{\sqrt{q^{2}\left(\tan ^{2} \theta+1\right)}}{p}+\frac{q}{p}\right)^{2}$

$\Rightarrow\left(\frac{\sqrt{\mathrm{q}^{2} \sec ^{2} \theta}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}$ [∵ 1+ tan² θ = sec² θ]

$\Rightarrow\left(\frac{\mathrm{q} \sec \theta}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}$

$\Rightarrow \frac{\mathrm{q}^{2}(\sec \theta+1)^{2}}{\mathrm{p}^{2}}$

$\Rightarrow \frac{\mathrm{q}^{2}(\sec \theta+1)^{2}}{\mathrm{q}^{2} \tan ^{2} \theta}$ (from Eq. (1))

$\Rightarrow \frac{(\sec \theta+1)^{2}}{\left(\sec ^{2} \theta-1\right)}$

$\Rightarrow \frac{(\sec \theta+1)(\sec \theta+1)}{(\sec \theta-1)(\sec \theta+1)}$ [∵(a + b) (a – b) = (a² – b²)]

$\Rightarrow \frac{\sec \theta+1}{\sec \theta-1}$

Now, we solve the RHS

$=\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}+\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}-\mathrm{q}}$

$=\frac{\sqrt{\mathrm{q}^{2} \tan ^{2} \theta+\mathrm{q}^{2}}+\mathrm{q}}{\sqrt{\mathrm{q}^{2} \tan ^{2} \theta+\mathrm{q}^{2}}-\mathrm{q}}$

$=\frac{\sqrt{\mathrm{q}^{2}\left(\tan ^{2} \theta+1\right)}+\mathrm{q}}{\sqrt{\mathrm{q}^{2}\left(\tan ^{2} \theta+1\right)}-\mathrm{q}}$

$=\frac{\sqrt{\mathrm{q}^{2} \sec ^{2}}+\mathrm{q}}{\sqrt{\mathrm{q}^{2} \sec ^{2} \theta}-\mathrm{q}}$ [∵ 1+ tan² θ = sec² θ]

$=\frac{\operatorname{qsec} \theta+q}{\operatorname{qsec} \theta-q}$

$\Rightarrow \frac{\sec \theta+1}{\sec \theta-1}$

∴ LHS = RHS
Hence Proved
Question 45 In the given figure, BC = 15 cm and sin B = 4/5, show that $\tan ^{2} \mathrm{B}-\frac{1}{\cos ^{2} \mathrm{B}}=-1$

Sol :
Given: BC =15cm and $\sin B=\frac{4}{5}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{B}=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{4}{5}$
Let,
Side opposite to angle B = 4k
and Hypotenuse = 5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AC)² + (BC)² = (AB)²
⇒ (4k)² + (BC)² = (5)²
⇒ 16k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –16 k²
⇒ (BC)² = 9 k²
⇒ BC =√9 k²
⇒ BC =±3k
But side BC can’t be negative. So, BC = 3k
Now, we have to find the value of cos B and tan B

We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle B = BC =3k
Hypotenuse = AB =5k
So, $\cos B=\frac{3 k}{5 k}=\frac{3}{5}$
Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
side opposite to angle B = AC =4k
Side adjacent to angle B = BC =3k
So, $\tan \mathrm{B}=\frac{4 \mathrm{k}}{3 \mathrm{k}}=\frac{4}{3}$

Now, $\tan ^{2} \mathrm{B}-\frac{1}{\cos ^{2} \mathrm{B}}$

$=\left(\frac{4}{3}\right)^{2}-\left(\frac{1}{\frac{3}{5}}\right)^{2}$

$=\frac{16}{9}-\frac{25}{9}$

$=\frac{-9}{9}$

= –1 = RHS
Hence Proved

Question 46 

In the given figure, find 3 tan θ – 2 sin α + 4 cos α.

Sol :
First of all, we find the value of RS
In right angled ∆RQS, we have
(RQ)² + (QS)² = (RS)²
⇒ (8)² + (6)² = (RS)²
⇒ 64 + 36 = (RS)²
⇒ RS =√100
⇒ RS =±10 [taking positive square root, since side cannot be negative]
⇒ RS =10

$\therefore \sin \alpha=\frac{P}{H}=\frac{8}{10}=\frac{4}{5}$

$\cos \alpha=\frac{B}{H}=\frac{6}{10}=\frac{3}{5}$

Now, we find the value of QP
In right angled ∆RQP
(RQ)² + (QP)² = (RP)²
⇒ (8)² + (QP)² = (17)²
⇒ 64 + (QP)² = 289
⇒ (QP)² =289–64
⇒ (QP)² =225
⇒ QP =√225
⇒ QP =±15 [taking positive square root, since side cannot be negative]
⇒ QP =15
tan θ$=\frac{\mathrm{P}}{\mathrm{B}}=\frac{8}{15}$

Now, 3 tan θ – 2 sinα + 4cos α
⇒ $3\left(\frac{8}{15}\right)-2\left(\frac{4}{5}\right)+4\left(\frac{3}{5}\right)$
⇒ $\frac{24}{15}-\frac{8}{5}+\frac{12}{5}$
⇒ $\frac{24-24+36}{15}$

⇒$\frac{36}{15}$

⇒$\frac{12}{5}$

Question 47 

In the given figure ΔABC is right angled at B and BD is perpendicular to AC. Find (i) cos θ, (ii) cot α.

Sol :
Firstly, we find the value of AC
In right angled ∆ABC
(AB)² + (BC)² = (AC)²
⇒ (12)² + (5)² = (AC)²
⇒ 144+25 =(AC)²
⇒ (AC)² =169
⇒ AC =√169
⇒ AC =±13
⇒ AC =13 [taking positive square root since, side cannot be negative]
(i) $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{12}{13}$
(ii) $\cot \alpha=\frac{\text { Base }}{\text { Perpendicular }}=\frac{12}{5}$

Question  48 

If 5 sin² θ + cos² θ = 2, find the value of sin θ.Sol :
Given: 5 sin² θ + cos² θ = 2
⇒ 5 sin² θ + (1– sin² θ) = 2 [∵ sin² θ + cos² θ = 1]
⇒ 5 sin² θ + 1 – sin² θ = 2
⇒ 4 sin² θ = 2 – 1
⇒ 4 sin² θ = 1

$\Rightarrow \quad \sin ^{2} \theta=\frac{1}{4}$

$\Rightarrow \quad \sin \theta=\sqrt{\frac{1}{4}}$

$\Rightarrow \quad \sin \theta=\pm \frac{1}{2}$

Question 49 

If 7 sin² θ + 3 cos² θ =4, find the value of tan θ.Sol :
Given : 7 sin² θ + 3 cos² θ =4
⇒ 7 sin² θ + 3(1– sin² θ) = 4 [∵ sin² θ + cos² θ = 1]
⇒ 7 sin² θ + 3 –3 sin² θ = 4
⇒ 4 sin² θ = 4 – 3
⇒ 4 sin² θ = 1

$\Rightarrow \sin ^{2} \theta=\frac{1}{4}$

$\Rightarrow \sin \theta=\sqrt{\frac{1}{4}}$

$\Rightarrow \sin \theta=\pm \frac{1}{2}$

Put the value of $\sin ^{2} \theta=\frac{1}{4}$ in given equation, we get

$\Rightarrow 7\left(\frac{1}{2}\right)^{2}+3 \cos ^{2} \theta=4$

$\Rightarrow \frac{7}{4}+3 \cos ^{2} \theta=4$

$\Rightarrow 3 \cos ^{2} \theta=4-\frac{7}{4}$

$\Rightarrow 3 \cos ^{2} \theta=\frac{16-7}{4}$

$\Rightarrow 3 \cos ^{2} \theta=\frac{9}{4}$

$\Rightarrow \cos ^{2} \theta=\frac{3}{4}$

$\Rightarrow \cos \theta=\sqrt{\frac{3}{4}}$

$\Rightarrow \cos \theta=\pm \frac{\sqrt{3}}{2}$

Now, we know that tan θ $=\frac{\sin \theta}{\cos \theta}$

$\Rightarrow \tan \theta=\frac{\left(\pm \frac{1}{2}\right)}{\left(\pm \frac{\sqrt{3}}{2}\right)}$

$\Rightarrow \tan \theta=\pm \frac{1}{\sqrt{3}}$

Question 50 

If 4 cos θ + 3 sin θ = 5, find the value of tan θ.Sol :
Given : 4 cos θ+ 3 sin θ = 5
Squaring both the sides, we get
⇒ (4 cos θ+ 3 sin θ)² = 25
⇒ 16 cos² θ + 9 sin² θ + 2(4cos θ)(3sin θ)= 25 [∵ (a + b)² =a² +b² +2ab]
⇒ 16 cos² θ + 9 sin² θ + 24 cosθ sinθ = 25

Divide by cos² θ, we get
$\Rightarrow \frac{16 \cos ^{2} \theta}{\cos ^{2} \theta}+\frac{9 \sin ^{2} \theta}{\cos ^{2} \theta}+\frac{24 \cos \theta \sin \theta}{\cos ^{2} \theta}=\frac{25}{\cos ^{2} \theta}$
⇒ 16 + 9tan² θ + 24 tanθ = 25sec² θ
⇒ 16 + 9tan² θ + 24 tanθ = 25(1 + tan² θ) [∵ 1+ tan²θ = sec² θ]
⇒ 16 + 9tan² θ + 24 tanθ = 25+ 25 tan² θ
⇒ 16tan² θ – 24tanθ + 9 = 0
⇒ 16tan² θ – 12 tanθ – 12 tanθ +9 = 0
⇒ 4tanθ (4tan θ – 3) – 3(4tan θ – 3) = 0
⇒ (4tan θ – 3)² = 0
$\Rightarrow \tan \theta=\frac{3}{4}$

Question 51 

If 7 sin A + 24 cos A = 25, find the value of tan A.Sol :
Given : 7 sin A + 24 cos A = 25
Squaring both the sides, we get
⇒ (7 sin A + 24 cos A)² = 625
⇒ 49 sin² A +576 cos² A + 2(7sin A) (24cos A) = 625 [∵ (a + b)² =a² +b² +2ab]
⇒ 49 sin² A +576 cos² A + 336 cosA sinA = 625

Divide by cos² θ, we get
$\Rightarrow \frac{49 \sin ^{2} \mathrm{A}}{\cos ^{2} \mathrm{A}}+\frac{576 \cos ^{2} \mathrm{A}}{\cos ^{2} \mathrm{A}}+\frac{336 \cos \mathrm{A} \sin \mathrm{A}}{\cos ^{2} \mathrm{A}}=\frac{625}{\cos ^{2} \mathrm{A}}$

⇒ 49tan² A +576+ 336 tanA = 625sec² A
⇒ 49tan² A +576+ 336 tanA = 625(1 + tan² A) [∵ 1+ tan²θ = sec² θ]
⇒ 49tan² A +576+ 336 tanθA = 625+625 tan² A
⇒ 576tan² A – 336tanA + 49 = 0
⇒ 576tan² A – 168 tanA – 168 tanA +49 = 0
⇒ 24tanθ (24tan A – 7) – 7(24tan A – 7) = 0
⇒ (24tan A – 7)² = 0
$\Rightarrow \tan \mathrm{A}=\frac{7}{24}$

Question 52 

If 9 sin θ + 40 cos θ= 41, find the value of cos θ and cosec θSol :
Given: 9 sin θ + 40 cos θ= 41
⇒ 9sinθ = 41 – 40 cosθ …(i)
Squaring both sides, we get
⇒ 81sin² θ = 1681+1600 cos² θ – 2(41) (40cos θ) [∵ (a – b)² =a² +b² –2ab]
⇒ 81 (1– cos² θ) =1681+1600 cos² θ – 3280cosθ
⇒ 81 – 81cos² θ = 1681 +1600cos² θ – 3280 cosθ
⇒ 1681cos² θ –3280cos θ +1600 = 0
⇒ (41)² cos² θ – 2(41) (40cos θ) + (40)² = 0
⇒ (41cos θ – 40 )² = 0
$\Rightarrow \cos \theta=\frac{40}{41}$Now, putting the value of cos θ in Eq. (i), we get

$\Rightarrow 9 \sin \theta=41-40\left(\frac{40}{41}\right)$

$\Rightarrow 9 \sin \theta=\left(\frac{1681-1600}{41}\right)$

$\Rightarrow \sin \theta=\left(\frac{81}{41 \times 9}\right)$

$\Rightarrow \frac{1}{\operatorname{cosec} \theta}=\left(\frac{9}{41}\right)$

$\Rightarrow \operatorname{cosec} \theta=\frac{41}{9}$

Question 53 

If tan A + sec A = 3, find the value of sin A.Sol :
tan A + sec A = 3
⇒ tanA = 3 – secA
Squaring both the sides, we get
⇒ tan² A =(3 – secA)²
⇒ tan² A = 9 + sec²A – 6sec A
⇒ sec² A – 1 = 9 + sec²A – 6sec A [∵ 1+ tan² A = sec² A]
⇒ –1 – 9 = –6secA
⇒ – 10 = –6sec A
$\Rightarrow \sec A=\frac{10}{6}$
$\Rightarrow \frac{1}{\cos A}=\frac{5}{3}\left[\because \sec A=\frac{1}{\cos A}\right]$
$\Rightarrow \cos A=\frac{3}{5}$

Now, tan A + sec A = 3
$\Rightarrow \frac{\sin A}{\cos A}+\frac{1}{\cos A}=3\left[\because \tan A=\frac{\sin A}{\cos A}\right]$
$\Rightarrow \frac{\sin A}{\cos A}=\frac{3 \cos A-1}{\cos A}$
⇒ sin A = 3cosA – 1

$\Rightarrow \sin A=3\left(\frac{3}{5}\right)-1$

$\Rightarrow \sin A=\left(\frac{9-5}{5}\right)$

$\Rightarrow \sin A=\left(\frac{4}{5}\right)$

Question 54 

If cosec A + cot A = 5, find the value of cos A.Sol :
cosec A + cot A = 5
⇒ cotA = 5 – cosecA
Squaring both the sides, we get
⇒ cot² A =(5 – cosecA)²
⇒ cot² A = 25 + cosec²A – 10cosec A
⇒ cosec² A – 1 = 25 + cosec²A – 10cosec A [∵ 1+ cot² A = cosec² A]
⇒ –1 – 25 = –10cosecA
⇒ – 26 = –10cosec A
$\Rightarrow \operatorname{cosec} A=\frac{26}{10}$
$\Rightarrow \frac{1}{\sin A}=\frac{13}{5}\left[\because \operatorname{cosec} A=\frac{1}{\sin A}\right]$
$\Rightarrow \sin \mathrm{A}=\frac{5}{13}$

Now, cosec A + cot A = 5
$\Rightarrow \frac{1}{\sin A}+\frac{\cos A}{\sin A}=5\left[\because \cot A=\frac{\cos A}{\sin A}\right]$
$\Rightarrow \frac{13}{5}+\frac{\cos A}{\frac{5}{13}}=5$

$\Rightarrow \frac{13}{5}+\frac{13 \cos A}{5}=5$

$\Rightarrow \frac{13 \cos A}{5}=5-\frac{13}{5}$

$\Rightarrow \frac{13 \cos A}{5}=\frac{25-13}{5}$

$\Rightarrow \cos A=\frac{12}{13}$

Question 55 

If tan θ + sec θ = x, show that $\sin \theta=\frac{x^{2}-1}{x^{2}+1}$Sol :
tan θ+ sec θ = x
⇒ tan θ = x – sec θ
Squaring both sides, we get
⇒ tan² θ =(x – secθ)²
⇒ tan² θ = x² + sec²θ – 2xsec θ
⇒ sec² θ – 1 = x² + sec²θ – 2xsec θ [∵ 1+ tan² A = sec² A]
⇒ –1 – x² = –2xsecθ
$\Rightarrow \sec \theta=\frac{1+x^{2}}{2 x}$
Now,
tan θ = x – sec θ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=x-\sec \theta$
$\Rightarrow \sin \theta\left(\frac{1+x^{2}}{2 x}\right)=x-\left(\frac{1+x^{2}}{2 x}\right)$
$\Rightarrow \sin \theta\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)=\left(\frac{2 \mathrm{x}^{2}-1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)$
$\Rightarrow \sin \theta\left(\frac{1+x^{2}}{2 x}\right)=\left(\frac{x^{2}-1}{2 x}\right)$
$\Rightarrow \sin \theta=\frac{x^{2}-1}{x^{2}+1}$ = RHS
Hence Proved

Question 56 

If cos θ +sin θ=1, prove that cos θ – sin θ = ± 1Sol :
Using the formula,
(a+b)² + (a – b)² = 2(a²+b²)
⇒ (cos θ +sin θ)² + (cos θ – sin θ)² = 2(cos²θ + sin² θ)
⇒ 1 + (cos θ – sin θ)² = 2(1)
⇒ (cos θ – sin θ)² = 2 –1
⇒ (cos θ – sin θ)² = 1
⇒ (cos θ – sin θ) =√1
⇒ (cos θ – sin θ) = ±1