KC Sinha: Exercise 4.3 – Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities

Question 1 

Express the following as trigonometric ratio of complementary angle of θ.

(i) cos θ (ii) sec θ
(iii) cot θ (iv) cosec θ
(v) tan θ
Sol :

(i) We know that
$\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}=\frac{A B}{A C}=\sin \left(90^{\circ}-\theta\right)$
⇒ cosθ = sin (90° – θ)

(ii) We know that
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{A B}=\operatorname{cosec}\left(90^{\circ}-\theta\right)$

(iii) We know that
$\cot \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\tan \left(90^{\circ}-\theta\right)$

(iv) We know that
$\operatorname{cosec} \theta=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\sec \left(90^{\circ}-\theta\right)$

(v) We know that
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{\mathrm{BC}}{\mathrm{AB}}=\cot \left(90^{\circ}-\theta\right)$

Question 2 

Express the following as trigonometric ratio of complementary angle of 90^o– θ.

(i) tan (90^o– θ)
(ii) cos (90^o– θ)
Sol :
(i) We know that,
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \tan \left(90^{\circ}-\theta\right)=\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \left(90^{\circ}-\theta\right)}$
$\Rightarrow \tan \left(90^{\circ}-\theta\right)=\frac{\sin 90^{\circ} \cos \theta-\cos 90^{\circ} \sin \theta}{\cos 90^{\circ} \cos \theta+\sin 90^{\circ} \sin \theta}$[∵ sin 90° = 1 and cos 90° = 0]
$\Rightarrow \tan \left(90^{\circ}-\theta\right)=\frac{\cos \theta}{\sin \theta}$
⇒ tan (90° – θ ) = cot θ

(ii) We know that.
Cos(A – B) = cos A cos B + sin A sin B
⇒ cos (90^o– θ) = cos 90° cos θ + sin 90° sin θ
⇒ cos (90^o– θ) = (0) cos θ + (1) sin θ
⇒ cos (90^o– θ) = sin θ

Question 3 

Fill up the blanks by an angle between 0^oand 90^o:

(i) sin 70^o = cos(…) (ii) sin 35^o = cos(…)
(iii) cos 48^o=sin (…) (iv) cos 70^o = sin (…)
(v) cos 50^o = sin (…) (vi) sec 32^o=cosec(…)
Sol :
(i) We know that
Sin θ = cos (90° – θ)
Here, θ = 70°
⇒ sin 70° = cos(90° -70°)
⇒ sin 70° = cos 20°

(ii) We know that
Sin θ = cos (90° – θ)
Here, θ = 35°
⇒ sin 35° = cos(90° -35°)
⇒ sin 35° = cos 55°

(iii) cos θ = sin (90° – θ)
Here, θ = 48°
⇒ cos 48° = sin (90° – 48°)
⇒ cos 48° = sin 42°

(iv) cos θ = sin (90° – θ)
Here, θ = 70°
⇒ cos 70° = sin (90° – 70°)
⇒ cos 70° = sin 20°

(v) cos θ = sin (90° – θ)
Here, θ = 50°
⇒ cos 50° = sin (90° – 50°)
⇒ cos 50° = sin 40°

(vi) sec θ = cosec (90°-θ)
Here, θ = 32°
⇒ sec 32° = cosec(90° – 32°)
⇒ sec 32° = cosec 58°

Question 4 

If A+B=90^o, then fill up the blanks with suitable trigonometric ratio of complementary angle of A or B.

(i) sin A =…. (ii) cos B =…
(iii) sec A =… (iv) tan B =…
(v) cosec B =… (vi) cot A=…
Sol :
(i) Here, A+B = 90°
⇒ A = 90° – B
Multiplying both sides by Sin, we get
Sin A = Sin (90° – B)
⇒ sin A = Cos B [∵ cos θ = sin (90° – θ)]
(ii) Here, A+B = 90°
⇒ B = 90° – A
Multiplying both sides by cos, we get
Cos B = cos (90° – A)
⇒ cos B = sin A [∵ Sin θ = cos (90° – θ)]
(iii) Here, A+B = 90°
⇒ A = 90° – B
Multiplying both sides by sec, we get
Sec A = Sec (90° – B)
⇒ sec A = Cosec B [∵ cosec θ = sec (90° – θ)]
(iv) Here, A+B = 90°
⇒ B = 90° – A
Multiplying both sides by tan, we get
tan B = tan (90° – A)
⇒ tan B = cot A [∵ cot θ = tan (90° – θ)]
(v) Here, A+B = 90°
⇒ B = 90° – A
Multiplying both sides by cosec, we get
Cosec B = cosec (90° – A)
⇒ cosec B = sec A [∵ sec θ = cosec (90° – θ)]
(vi) Here, A+B = 90°
⇒ A = 90° – B
Multiplying both sides by Sin, we get
cotA = cot (90° – B)
⇒ cot A = tan B [∵ tan θ = cot (90° – θ)]

Question 5 A 

If sin 37^o=a, then express cos 53^o in terms of a.

Sol :
Given sin 37° = a
We know that sin θ = cos (90° – θ)
Here, θ = 37°
⇒ cos (90° – 37°) = a
⇒ cos 53° = a

Question 5 B 

If cos 47^o=a, then express sin 43^o in terms of a.

Sol :
Given cos 47° = a
We know that cos θ = sin (90° – θ)
Here, θ = 47°
⇒ sin (90° – 47°) = a
⇒ sin 43° = a

Question 5 C 

If sin 52^o=a, then express sin 38^o in terms of a.

Sol :
Given sin 52° = a
We know that sin θ = cos (90° – θ)
Here, θ = 52°
⇒ cos (90° – 52°) = a
⇒ cos 38° = a

Question 5 D 

If sin 56^o=x, then express sin 34^o in terms of x.

Sol :
Given sin 56° = x
We know that sin θ = cos (90° – θ)
Here, θ = 56°
⇒ cos (90° – 56°) = x
⇒ cos 34° = x

Question 6 

Find the value of

(i) $\frac{\cos 59^{\circ}}{\sin 31^{\circ}}$ 

(ii) $\frac{\cos 53^{\circ}}{\sin 37^{\circ}}$

(iii) $\frac{\sin 20^{\circ}}{\cos 70^{\circ}}$

(iv) $\frac{\sqrt{2} \sin 22^{\circ}}{\cos 68^{\circ}}$

(v)$\frac{\sin 10^{\circ}}{\cos 80^{\circ}}$

(vi) $\frac{\sin 27^{\circ}}{\cos 63^{\circ}}$

(vii) $\frac{\sqrt{3} \cos 65^{\circ}}{\sin 25^{\circ}}$

(viii) $\frac{\cos 29^{\circ}}{\sin 61^{\circ}}$

(ix) sin 54° – cos 36° 

(x) $\frac{\tan 80^{\circ}}{\cot 10^{\circ}}$
(xi) cosec 31° – sec 59° 

(xii) $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$
(xiii) $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
Sol :
(i)$\frac{\cos 59^{\circ}}{\sin 31^{\circ}}=\frac{\sin \left(90^{\circ}-59^{\circ}\right)}{\sin 31^{\circ}}$ 

$=\frac{\sin 31^{\circ}}{\sin 31^{\circ}}=1$ [∵ cos θ = sin (90° – θ)]

(ii) $\frac{\cos 53^{\circ}}{\sin 37^{\circ}}=\frac{\sin \left(90^{\circ}-53^{\circ}\right)}{\sin 37^{\circ}}$

$=\frac{\sin 37^{\circ}}{\sin 37^{\circ}}=1$ [∵ cos θ = sin (90° – θ)]

(iii) $\frac{\sin 20^{\circ}}{\cos 70^{\circ}}=\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\cos 70^{\circ}}$

$=\frac{\cos 70^{\circ}}{\cos 70^{\circ}}=1$ [∵ Sin θ = cos (90° – θ)]

(iv) $\frac{\sqrt{2} \sin 22^{\circ}}{\cos 68^{\circ}}=\frac{\sqrt{2} \cos \left(90^{\circ}-22^{\circ}\right)}{\cos 68^{\circ}}$

$=\frac{\sqrt{2} \cos 68^{\circ}}{\cos 68^{\circ}}=\sqrt{2}$ [∵ Sin θ = cos (90° – θ)]

(v) $\frac{\sin 10^{\circ}}{\cos 80^{\circ}}=\frac{\cos \left(90^{\circ}-10^{\circ}\right)}{\cos 80^{\circ}}$

$=\frac{\cos 80^{\circ}}{\cos 80^{\circ}}=1$ [∵ Sin θ = cos (90° – θ)]

(vi) $\frac{\sin 27^{\circ}}{\cos 63^{\circ}}=\frac{\cos \left(90^{\circ}-27^{\circ}\right)}{\cos 63^{\circ}}$

$=\frac{\cos 63^{\circ}}{\cos 63^{\circ}}=1$ [∵ Sin θ = cos (90° – θ)]

(vii) $\frac{\sqrt{3} \cos 65^{\circ}}{\sin 25^{\circ}}=\frac{\sqrt{3} \sin \left(90^{\circ}-65^{\circ}\right)}{\sin 25^{\circ}}$

$=\frac{\sin 25^{\circ}}{\sin 25^{\circ}}=\sqrt{3}$ [∵ cos θ = sin (90° – θ)]

(viii) $\frac{\cos 29^{\circ}}{\sin 61^{\circ}}=\frac{\sin \left(90^{\circ}-29^{\circ}\right)}{\sin 61^{\circ}}$

$=\frac{\sin 61^{\circ}}{\sin 61^{\circ}}=1$ [∵ cos θ = sin (90° – θ)]

(ix) sin 54° – sin(90° – 36°) [∵ cos θ = sin (90° – θ)]
⇒ sin 54° – sin 54°
⇒ 0

(x) $\frac{\tan 80^{\circ}}{\cot 10^{\circ}}=\frac{\cot \left(90^{\circ}-80^{\circ}\right)}{\cot 10^{\circ}}$

$=\frac{\cot 10^{\circ}}{\cot 10^{\circ}}=1$ [∵ tan θ = cot (90° – θ)]

(xi) cosec 31° – cosec(90° – 59°) [∵ sec θ = cosec (90° – θ)]
⇒ cosec 31° – cosec 31°
⇒ 0

(xii) $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\frac{\cos \left(90^{\circ}-18^{\circ}\right)}{\cos 72^{\circ}}$

$=\frac{\cos 72^{\circ}}{\cos 72^{\circ}}=1$ [∵ Sin θ = cos (90° – θ)]

(xiii) $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=\frac{\cot \left(90^{\circ}-65^{\circ}\right)}{\cot 25^{\circ}}$

$=\frac{\cot 25^{\circ}}{\cot 25^{\circ}}=1$ [∵ tan θ = cot (90° – θ)]

Question 7 

Fill up the blanks :

(i) If sin 50^o=0.7660, then cos 40^o=……
(ii) If cos 44^o = 0.7193, then sin 46^o=…..
(iii) sin 50^o+cos 40^o = 2 sin (………)
(iv) Value of $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}$ is ………
Sol :
(i) Given: sin 50^o=0.7660
We know that
Sin θ = cos (90° – θ)
⇒ cos (90° – 50°) = 0.7660
⇒ cos 40° = 0.7660

(ii) Given: cos 44° = 0.7193
We know that,
cos θ = sin (90° – θ)
⇒ sin (90° – 44°) = 0.7193
⇒ sin 46° = 0.7193

(iii) LHS = sin 50° + cos 40°
⇒ sin 50° + sin (90° – 40°) [∵ cos θ = sin (90° – θ)]
⇒ sin 50° + sin 50°
⇒ 2sin 50°

(iii) $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}=\frac{\cos \left(90^{\circ}-70^{\circ}\right)}{\cos 20^{\circ}}$

$=\frac{\cos 20^{\circ}}{\cos 20^{\circ}}=1$ [∵ Sin θ = cos (90° – θ)]

Question 8 A 

If A + B = 90^o, then express cos B in terms of simplest trigonometric ratio of A.

Sol :
Given: A+B =90°
⇒ B = 90° – A
Multiplying both side by cos, we get
= cos B = Cos (90° – A)
⇒ cos B = sin A [∵ Sin θ = cos (90° – θ)]

Question 8 B 

If X + Y = 90^o, then express cos X in terms of simplest trigonometric ratio of Y.

Sol :
Given: X+Y =90°
⇒ X= 90° – Y
Multiplying both side by cos, we get
= cos X = Cos (90° – Y)
⇒ cos X = sin Y [∵ Sin θ = cos (90° – θ)]

Question 9 A 

If A + B = 90^o, sin A = a, sin B = b, then prove that

(a) a² + b² = 1
(b) $\tan A=\frac{a}{b}$
Sol :
(a) LHS = a² +b²
= (sin A)² + (sin B)²
= sin² A + sin² B
= sin² A + sin² (90° – A) [∵ cos θ = sin (90° – θ)]
= sin² A + cos² A
= 1 [∵ sin² θ + cos² θ = 1]
=RHS
Hence Proved

(b) LHS = tan A
Now, taking RHS $=\frac{a}{b}$
⇒ $\Rightarrow \frac{\sin A}{\sin B}$
$\Rightarrow \frac{\sin A}{\sin \left(90^{\circ}-A\right)}$ {given, A +B = 90°)
$\Rightarrow \frac{\sin A}{\cos A}$ [∵ cos θ = sin (90° – θ)]
⇒ tan A
=LHS
∴ LHS = RHS
Hence Proved

Question 9 B 

Show that sin(50° + θ) – cos (40° – θ) = 0.

Sol :
LHS = sin (50° + θ) – cos (40° – θ)
We know that,
Sin A = cos (90° – A)
Here, A = 50° + θ
⇒ cos {90° -( 50° + θ)} – cos (40° – θ)
⇒ cos (40° – θ) – cos (40° – θ)
= 0 = RHS
Hence Proved

Question 10 

Prove that $\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}=2$

Sol :
Taking LHS,
$\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}$ [∵ cos θ = sin (90° – θ) and Sin θ = cos (90° – θ)]
$\Rightarrow \frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\sin \theta}$
⇒ 1+ 1
= 2 = RHS
Hence Proved

Question 11 A 

In a ∆ABC prove that

$\sin \frac{\mathrm{B}+\mathrm{C}}{2}=\cos \frac{\mathrm{A}}{2}$
Sol :

In ∆ABC,
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ B + C = 180° – A

Multiplying both sides by $\frac{1}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}}{2}-\frac{\mathrm{A}}{2}$
$=\frac{B+C}{2}=90^{\circ}-\frac{A}{2}$ …(1)

Taking LHS
$\sin \frac{\mathrm{B}+\mathrm{C}}{2}$
$=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$ (from eq (1))
$=\cos \frac{A}{2}$ [∵ sin (90° – θ) = cos θ]
=RHS
Hence Proved

Question 11 B 

In a ∆ABC prove that

$\tan \frac{\mathrm{B}+\mathrm{C}}{2}=\cot \frac{\mathrm{A}}{2}$
Sol :

In ∆ABC,
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ B + C = 180° – A

Multiplying both sides by $\frac{1}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}}{2}-\frac{\mathrm{A}}{2}$
$=\frac{B+C}{2}=90^{\circ}-\frac{A}{2}$ …(2)

Taking LHS
$\tan \frac{\mathrm{B}+\mathrm{C}}{2}$
$=\tan \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$ (from eq (2))
$=\cot \frac{A}{2}$ [∵ tan (90° – θ) = cot θ]
=RHS
Hence Proved

Question 11 C 

In a ∆ABC prove that

$\cos \frac{A+B}{2}=\sin \frac{C}{2}$
Sol :

In ∆ABC,
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ A + B = 180° – C

Multiplying both sides by $\frac{1}{2}$
$=\frac{A+B}{2}=\frac{180^{\circ}-C}{2}$
$=\frac{A+B}{2}=\frac{180^{\circ}}{2}-\frac{C}{2}$
$=\frac{A+B}{2}=90^{\circ}-\frac{C}{2}$ …(3)

Taking LHS
$\cos \frac{A+B}{2}$
$=\cos \left(90^{\circ}-\frac{\mathrm{C}}{2}\right)$ (from eq (3))
$=\sin \frac{C}{2}$ [∵ cos (90° – θ) = sin θ]
=RHS
Hence Proved

Question 12 A 

If sin 3A = cos(A – 26^o), where 3A is an acute angle, find the value of A.

Sol :
sin 3A = cos (A-26°) …(i)
We know that
Sin θ = cos (90° – θ)
So, Eq. (i) become
Cos (90° – 3A) = cos (A -26°)
On Equating both the sides, we get
90° – 3A = A – 26°
⇒ -3A – A = -26° -90°
⇒ -4A = -116°
⇒ A = 29°

Question 12 B 

Find θ if cos(2 θ +54^o)= sin θ, where (2θ +54^o) is an acute angle.

Sol :
cos(2 θ +54^o)= sin θ …(i)
We know that
Sin θ = cos (90° – θ)
So, Eq. (i) become
cos(2 θ +54^o) = cos( 90° – θ)
On Equating both the sides, we get
2θ + 54° = 90° – θ
⇒ 2θ + θ = 90° – 54°
⇒ 3θ = 36°
⇒ θ = 12°

Question 12 C 

If tan 3 θ =cot (θ +18^o), where 3 θ and θ +18^o are acute angles, find the value of θ.

Sol :
tan 3θ = cot (θ + 18°) …(i)
We know that
tan θ = cot (90° – θ)
So, Eq. (i) become
Cot (90° – 3θ) = cot (θ + 18°)
On Equating both the sides, we get
90° – 3θ = θ + 18°
⇒ -3θ – θ = 18° -90°
⇒ -4θ = -72°
⇒ θ = 18°

Question 12 D 

If sec 5 θ =cosec (θ -36^o), where 5 θ is an acute angle, find the value of θ.

Sol :
sec 5θ = cosec (θ-36°) …(i)
We know that
sec θ = cosec (90° – θ)
So, Eq. (i) become
Cosec (90° – 5θ) = cosec (θ -36°)
On Equating both the sides, we get
90° – 5θ = θ -36°
⇒ -5θ – θ = -36° -90°
⇒ -6θ = -126°
⇒ θ = 21°

Question 13 

Prove that :

sin 70^o. sec 20^o=1
Sol :
Taking LHS
sin 70° sec 20°
$\Rightarrow \sin 70^{\circ} \times \frac{1}{\cos 20^{\circ}}$
$\Rightarrow \sin 70^{\circ} \times \frac{1}{\sin \left(90^{\circ}-20^{\circ}\right)}$ [∵ cos θ = sin (90° – θ)]
$\Rightarrow \frac{\sin 70^{\circ}}{\sin 70^{\circ}}$
= 1 = RHS
Hence Proved

Question 14 

Prove that :

sin (90^o– θ) tan θ=sin θ
Sol :
Taking LHS
Sin(90° – θ) tanθ [∵ cos θ = sin (90° – θ)]
⇒ cos θ tan θ
$\Rightarrow \cos \theta \times \frac{\sin \theta}{\cos \theta}$ [∵ tan θ $=\frac{\sin \theta}{\cos \theta}$]
= sin θ = RHS
Hence Proved

Question 15 

Prove that :

tan 63^o. tan 27^o=1
Sol :
Taking LHS
Tan 63° tan 27°
⇒ tan 63° cot (90° – 27°) [∵ tan θ = cot (90° – θ)]
⇒ tan 63° cot 63°

$\Rightarrow \tan 63^{\circ} \times \frac{1}{\tan 63^{\circ}}$ $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
= 1 =RHS
Hence Proved

Question 16

Prove that :

$\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1=-\sin ^{2} \theta$
Sol :
Taking LHS
$=\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$
$=\frac{\cos \theta \sin \theta}{\frac{\sin \theta}{\cos \theta}}-1$
$=\frac{\cos \theta \sin \theta \times \cos \theta}{\sin \theta}-1$
= cos² θ – 1
= – sin² θ [∵ cos² θ + sin² θ = 1]
= RHS
Hence Proved

Question 17 

Prove that :

sin 55^o. cos 48^o=cos35^o. sin 42^o

Sol :
Taking LHS = sin 55 ° cos 48°
We know that
cos θ = sin (90° – θ)
Here, θ = 48°
⇒ sin 55° sin (90° – 48°)
⇒ sin 55° sin 42°
We also know that
Sin θ = cos (90° – θ)
Here, θ = 55°
⇒ cos (90° – 55°) sin 42°
⇒ cos 35° sin 42° = RHS
Hence Proved

Question 18 

Prove that :

sin² 25^o+sin² 65° = cos² 63°+cos² 39^o
Sol :
Taking LHS = sin 25^o+sin65^o
We know that
Sin θ = cos (90° – θ)
Here, θ = 25°
⇒ cos² (90° – 25°)+ sin² 65°
⇒ cos² 65° + sin² 65°
= 1 [∵ cos² θ + sin² θ = 1]
Now, RHS = cos² 63^o+cos² 39^o
We know that
cos θ = sin (90° – θ)
Here, θ = 39°
⇒ cos² 63° + sin² (90° – 39°)
⇒ cos² 63°+ sin² 63°
=1 [∵ cos² θ + sin² θ = 1]
LHS = RHS
Hence Proved

Question 19 

Prove that :

sin 54^o+cos67^o= sin23^o+cos36^o
Sol :
Taking LHS = sin 54^o+cos67^o
We know that
cos θ = sin (90° – θ)
Here, θ = 67°
⇒ sin 54°+ sin (90° – 67°)
⇒ sin 54°+ sin 23°
We also know that
Sin θ = cos (90° – θ)
Here, θ = 54°
⇒ cos (90° – 54°)+ sin 23°
⇒ cos 36°+ sin 23° = RHS
Hence Proved

Question 20 

Prove that :

cos 27+ sin51^o = sin63^o+cos 39^o
Sol :
Taking LHS = cos 27+ sin51^o
We know that
cos θ = sin (90° – θ)
Here, θ = 27°
⇒ sin (90° – 27°)+ sin 51°
⇒ sin 63°+ sin 51°
We also know that
Sin θ = cos (90° – θ)
Here, θ = 51°
⇒ sin 63°+ cos (90° – 51°)
⇒ sin 63°+ cos 39° = RHS
Hence Proved

Question 21 

Prove that :

sin²40^o+sin²50^o=1
Sol :
Taking LHS= sin²40^o+sin²50^o
⇒ cos² (90° – 40°) + sin² 50° [∵ Sin θ = cos (90° – θ)]
⇒ cos² 50° + sin² 50°
= 1 =RHS [∵ cos² θ + sin² θ = 1]
Hence Proved

Question 22 

Prove that :

sin²29^o + sin²61^o=1
Sol :
Taking LHS= sin²29^o + sin²61^o
⇒ cos² (90° – 29°) + sin² 61° [∵ Sin θ = cos (90° – θ)]
⇒ cos² 61° + sin² 61°
= 1 =RHS [∵ cos² θ + sin² θ = 1]
Hence Proved

Question 23 

Prove that :

sin θ .cos (90° – θ) + cos θ sin (90° – θ).
Sol :
Taking LHS = sin θ cos (90° – θ) + cos θ sin (90° – θ)
⇒ sin θ × sin θ + cos θ × cos θ [∵ Sin θ = cos (90° – θ) and cos θ = sin (90° – θ) ]
⇒ cos² θ + sin² θ [∵ cos² θ + sin² θ = 1]
= 1 = RHS
Hence Proved

Question 24 

Prove that :

cos θ . cos(90° – θ) + sin θ sin (90° – θ) = 0
Sol :
Taking LHS = cos θ cos ( 90° – θ) + sin θ sin (90° – θ)
⇒ cos θ × sin θ – sinθ × cos θ [∵ Sin θ = cos (90° – θ) and cos θ = sin (90° – θ) ]
= 0 = RHS
Hence Proved

Question 25 

Prove that :

sin 42°. cos 48° + cos 42° . sin 48° = 1
Sol :
Taking LHS
= sin 42° cos 48° + cos 42° sin 48°
= cos (90° – 42°) cos 48° + sin (90° – 42°) sin 48°
[∵ Sin θ = cos (90° – θ) and cos θ = sin (90° – θ) ]
= cos 48° cos 48° + sin 48° sin 48°
= cos² 48° + sin² 48°
= 1 [∵ cos² θ + sin² θ = 1]
=LHS=RHS
Hence Proved

Question 26 

Prove that :

$\frac{\cos 20^{\circ}}{\sin 70^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}=2$
Sol :
Taking LHS
$=\frac{\cos 20^{\circ}}{\sin 70^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}$
$=\frac{\cos 20^{\circ}}{\cos \left(90^{\circ}-70^{\circ}\right)}+\frac{\cos \theta}{\cos \theta}$ [∵ Sin θ = cos (90° – θ) and cos θ = sin (90° – θ) ]
$=\frac{\cos 20^{\circ}}{\cos 20^{\circ}}+1$
= 1 + 1
= 2 = RHS
Hence Proved

Question 27 

Prove that :

tan 27° tan 45° tan 63°
Sol :
Taking LHS
= tan 27° tan 45° tan 63°
=tan (90° – 27°) tan 45° tan 63° [∵ tan θ = cot (90° – θ)]
=cot 63° tan 45° tan 63° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 63^{\circ}} \times \tan 45^{\circ} \times \tan 63^{\circ}$
= tan 45° [∵ tan 45° =1]
=1 =RHS
Hence Proved

Question 28 

Prove that :

tan 9°. tan 27°. tan 45°. tan 63°. tan 81° = 1
Sol :
Taking LHS
= tan 9° tan 27° tan 45° tan 63° tan 81°
=cot(90° – 9°) tan (90° – 27°) tan 45° tan 63° tan 81° [∵ tan θ = cot (90° – θ)]
=cot 81° cot 63° tan 45° tan 63° tan 81° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 81^{\circ}} \times \frac{1}{\tan 63^{\circ}} \times \tan 45^{\circ} \times \tan 63^{\circ} \times \tan 81^{\circ}$
= tan 45° [∵ tan 45° =1]
=1 =RHS
Hence Proved

Question 29 

Prove that :

sin 9°. sin 27°. sin 63°. sin 81°
= cos9°.cos27°.cos63°.cos81°
Sol :
Taking LHS
= sin 9° sin 27° sin 63° sin 81°
= cos (90° – 9°) cos (90° – 27°) cos (90° – 63°) cos (90°- 81°)
= cos 81° cos 63° cos 27° cos 9°
Or cos 9° cos 27° cos 63° cos 81° = RHS
Hence Proved

Question 30 A 

Prove that :

$\tan 7^{\circ} \cdot \tan 23^{\circ} \cdot \tan 60^{\circ} \cdot \tan 67^{\circ} \cdot \tan 83^{\circ}=\sqrt{3}$
Sol :
Taking LHS
= tan 7° tan 23° tan 60° tan 67° tan 83°
=cot(90° – 7°) tan (90° – 23°) tan 60° tan 67° tan 83° [∵ tan θ = cot (90° – θ)]
=cot 83° cot 67° tan 60° tan 67° tan 83° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 83^{\circ}} \times \frac{1}{\tan 67^{\circ}} \times \tan 60^{\circ} \times \tan 67^{\circ} \times \tan 83^{\circ}$
= tan 60° [∵ tan 60° =√3]
=√3 =RHS

Question 30 B 

Prove that :

$\tan 15^{\circ} \tan 25^{\circ} \tan 60^{\circ} \tan 65^{\circ} \tan 75^{\circ}=\sqrt{3}$
Sol :
Taking LHS
= tan 15° tan 25° tan 60° tan 65° tan 75°
=cot(90° – 15°) tan (90° – 25°) tan 60° tan 65° tan 75° [∵ tan θ = cot (90° – θ)]
=cot 75° cot 65° tan 60° tan 65° tan 75° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 75^{\circ}} \times \frac{1}{\tan 65^{\circ}} \times \tan 60^{\circ} \times \tan 65^{\circ} \times \tan 75^{\circ}$
= tan 60° [∵ tan 60° =√3]
=√3 =RHS

Question 31 

Find the value off the following:

$\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\cos \mathrm{ec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \cdot \cos \mathrm{ec} 40^{\circ}$
Sol :
$=\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}$

$=\frac{\sin 50^{\circ}}{\sin \left(90^{\circ}-40^{\circ}\right)}+\frac{\operatorname{cosec} 40^{\circ}}{\operatorname{cosec}\left(90^{\circ}-50^{\circ}\right)}-4 \sin \left(90^{\circ}-50^{\circ}\right) \operatorname{cosec} 40^{\circ}$
[∵ cos θ = sin (90° – θ) and sec θ = cosec (90° – θ)]

$=\frac{\sin 50^{\circ}}{\sin 50^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\operatorname{cosec} 40^{\circ}}-4 \sin 40^{\circ} \times \frac{1}{\sin 40^{\circ}}$ $\left[\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}\right]$

= 1 + 1 – 4
= -2

Question 32 

Find the value off the following:

$\frac{\cos ^{2} 20^{0}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}+\sin 35^{\circ} \cdot \sec 55^{\circ}$
Sol :
$=\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}+\cos \left(90^{\circ}-35^{\circ}\right) \sec 55^{\circ}$

$=\frac{\cos ^{2} 20^{\circ}+\sin ^{2}\left(90^{\circ}-70^{\circ}\right)}{\sin ^{2}\left(59^{\circ}\right)+\cos ^{2}\left(90^{\circ}-31^{\circ}\right)}+\cos 55^{\circ} \sec 55^{\circ}$
[∵ cos θ = sin (90° – θ) and sec θ = cosec (90° – θ)]

$=\frac{\cos ^{2} 20^{\circ}+\sin ^{2} 20^{\circ}}{\sin ^{2}\left(59^{\circ}\right)+\cos ^{2} 59^{\circ}}+\cos 55^{\circ} \times \frac{1}{\cos 55^{\circ}}$

= 1 + 1
=2

Question 33 

Find the value off the following:

$\frac{\tan 50^{\circ}+\sec 50^{\circ}}{\cot 40^{\circ}+\cos \mathrm{ec} 40^{\circ}}+\cos 40^{\circ} \cdot \operatorname{cosec} 50^{\circ}$
Sol :
$=\frac{\tan 50^{\circ}+\sec 50^{\circ}}{\cot 40^{\circ}+\operatorname{cosec} 40^{\circ}}+\cos 40^{\circ} \operatorname{cosec} 50^{\circ}$

$=\frac{\cot \left(90^{\circ}-50^{\circ}\right)+\operatorname{cosec}\left(90^{\circ}-50^{\circ}\right)}{\cot 40^{\circ}+\operatorname{cosec} 40^{\circ}}+\sin \left(90^{\circ}-40^{\circ}\right) \operatorname{cosec} 50^{\circ}$
[∵ tan θ = cot (90° – θ) , sec θ = cosec (90° – θ) and cos θ = sin (90° – θ)]

$=\frac{\cot 40^{\circ}+\operatorname{cosec} 40^{\circ}}{\cot 40^{\circ}+\operatorname{cosec} 40^{\circ}}+\sin 50^{\circ} \operatorname{cosec} 50^{\circ}$

$=1+\sin 50^{\circ} \times \frac{1}{\sin 50^{\circ}}$ $\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}$

= 1 + 1
= 2

Question 34 

Find the value off the following:

cosec (65° + θ) – sec (25° – θ) – tan(55° – θ) + cot(35° +θ)
Sol :
cosec (65° + θ) – sec (25° – θ) – tan(55° – θ) + cot(35° +θ)
= sec {90°-(65°+θ)} – sec (25° – θ) – tan(55° – θ) + tan {90°-(35° +θ)}
[∵ cosec θ = sec (90° – θ) and cot θ = tan (90° – θ)]
= sec ( 90° – 65°-θ) – sec (25° – θ) – tan(55° – θ) + tan (90°- 35° – θ)
= sec (25° – θ) – sec (25° – θ) – tan(55° – θ) + tan (55° – θ)
= 0

Question 35 

Find the value off the following:

$\frac{\cos 35^{\circ}}{\sin 55^{\circ}}+\frac{\sin 11^{\circ}}{\cos 79^{\circ}}-\cos 28^{\circ} \cdot \operatorname{cosec} 62^{\circ}$
Sol :
$=\frac{\cos 35^{\circ}}{\sin 55^{\circ}}+\frac{\sin 11^{\circ}}{\cos 79^{\circ}}-\cos 28^{\circ} \operatorname{cosec} 62^{\circ}$

$=\frac{\sin \left(90^{\circ}-35^{\circ}\right)}{\sin 55^{\circ}}+\frac{\sin 11^{\circ}}{\sin \left(90^{\circ}-79^{\circ}\right)}-\sin \left(90^{\circ}-28^{\circ}\right) \operatorname{cosec} 62^{\circ}$

$=\frac{\sin 55^{\circ}}{\sin 55^{\circ}}+\frac{\sin 11^{\circ}}{\sin 11^{\circ}}-\sin 62^{\circ} \operatorname{cosec} 62^{\circ}$

$=1+1-\sin 62^{\circ} \times \frac{1}{\sin 62^{\circ}}$
= 1 + 1 – 1
= 1

Question 36 

Find the value off the following:

$\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}$
Sol :
$=\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}$

$=\frac{\cos ^{2} 20^{\circ}+\sin ^{2}\left(90^{\circ}-70^{\circ}\right)}{\sin ^{2}\left(59^{\circ}\right)+\cos ^{2}\left(90^{\circ}-31^{\circ}\right)}$
[∵ cos θ = sin (90° – θ) and sec θ = cosec (90° – θ)]

$=\frac{\cos ^{2} 20^{\circ}+\sin ^{2} 20^{\circ}}{\sin ^{2}\left(59^{\circ}\right)+\cos ^{2} 59^{\circ}}$
= 1

Question 37 

Find the value off the following:

cosec (65° + θ) – sec (25° – θ)
Sol :
cosec (65° + θ) – sec (25° – θ)
= sec {90°-(65°+θ)} – sec (25° – θ)
[∵ cosec θ = sec (90° – θ)]
= sec ( 90° – 65°-θ) – sec (25° – θ)
= sec (25° – θ) – sec (25° – θ)
= 0

Question 38 

Find the value off the following:

cos (60° + θ) – sin (30° – θ)
Sol :
cos (60° + θ) – sin (30° – θ)
= sin {90°-(60°+θ)} – sin (30° – θ) [∵ cos θ = sin (90° – θ)]
= sin ( 90° – 60°-θ) – sin (30° – θ)
= sin (30° – θ) – sin (30° – θ)
= 0

Question 39 

Find the value off the following:

sec 70°. sin 20° – cos 20°. cosec 70°
Sol :
sec 70° sin 20° – cos 20° cosec 70°
= cosec (90°-70°) cos (90° – 20°)- cos 20° cosec 70°
[∵ sec θ = cosec (90° – θ) and Sin θ = cos (90° – θ)]
= cosec 70° cos 20° – cos 20° cosec 70°
=0

Question 40 

Find the value off the following:

(sin 72° + cos 18°)( sin 72° – cos 18°)
Sol :
(sin 72° + cos 18°)( sin 72° – cos 18°)
Using the identity , (a-b)(a+b) = a² – b²
= (sin 72°)² – (cos 18°)²
= {cos(90° – 72°)}² – (cos 18°)² [∵ Sin θ = cos (90° – θ)]
=(cos 18°)² – (cos 18°)²
= 0

Question 41 

Find the value off the following:

$\left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^{2}+\left(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\right)-2 \cos 60^{\circ}$
Sol :
$=\left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^{2}+\left(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\right)^{2}-2 \cos 60^{\circ}$
$=\left(\frac{\sin 35^{\circ}}{\sin \left(90^{\circ}-55^{\circ}\right)}\right)^{2}+\left(\frac{\sin \left(90^{\circ}-55^{\circ}\right)}{\sin 35^{\circ}}\right)^{2}-2 \cos 60^{\circ}$ [∵ cos θ = sin (90° – θ)]

$=\left(\frac{\sin 35^{\circ}}{\sin 35^{\circ}}\right)^{2}+\left(\frac{\sin 35^{\circ}}{\sin 35^{\circ}}\right)^{2}-2\left(\frac{1}{2}\right)$
= 1 + 1 – 1
= 1

Question 42 

Find the value off the following:

$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \cdot \operatorname{cosec} 31^{\circ}$
Sol :
$=\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$

$=\frac{\sin \left(90^{\circ}-80^{\circ}\right)}{\sin 10^{\circ}}+\sin \left(90^{\circ}-59^{\circ}\right) \operatorname{cosec} 31^{\circ}$

$=\frac{\sin 10^{\circ}}{\sin 10^{\circ}}+\sin 31^{\circ} \operatorname{cosec} 31^{\circ}$

$=1+\sin 31^{\circ} \times \frac{1}{\sin 31^{\circ}}$
= 1 + 1
= 2

Question 43 

Find the value off the following:

(sin 50° + θ)- cos (40° – θ) + tan 1°. tan 10° tan 20°. tan 70°. tan 80°. tan 89°
Sol :
(sin 50° + θ)- cos (40° -θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°

= cos {90° -(50° + θ)} – cos (40° -θ)+ cot(90° – 1°) tan (90° – 10°) cot(90° – 20°) tan 70° tan 80° tan 89° [∵ Sin θ = cos (90° – θ) & tan θ = cot (90° – θ)]

= cos (40° -θ) – cos (40° -θ) + cot89° cot80° cot70° tan 70° tan 80° tan 89°

$=\frac{1}{\tan 89^{\circ}} \times \frac{1}{\tan 80^{\circ}} \times \frac{1}{\tan 70^{\circ}} \times \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}$ $\left[\because \cot \theta=\frac{1}{\tan \theta}\right]$
= 1

Question 44 

Find the value off the following:

$\sec ^{2} 10^{\circ}-\cot ^{2} 80^{\circ}+\frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \cdot \sin 75^{\circ}}{\cos \theta \sin (90-\theta)+\sin \theta \cos \left(90^{\circ}-\theta\right)}$
Sol :
$=\sec ^{2} 10^{\circ}-\cot ^{2} 80^{\circ}+\frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \sin 75^{\circ}}{\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos \left(90^{\circ}-\theta\right)}$

$=\sec ^{2} 10^{\circ}-\tan ^{2}\left(90^{\circ}-80^{\circ}\right)+\frac{\cos \left(90^{\circ}-15^{\circ}\right) \cos 75^{\circ}+\sin \left(90^{\circ}-15^{\circ}\right) \sin 75^{\circ}}{\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos \left(90^{\circ}-\theta\right)}$

[∵ cot θ = tan (90° – θ), cos θ = sin (90° – θ) and Sin θ = cos (90° – θ)]

$=\sec ^{2} 10^{\circ}-\tan ^{2} 10^{\circ}+\frac{\cos 75^{\circ} \cos 75^{\circ}+\sin 75^{\circ} \sin 75^{\circ}}{\cos \theta \cos (\theta)+\sin \theta \sin \theta}$
[∵ 1+tan² θ =sec² θ and cos² θ +sin²θ = 1]

$=1+\frac{\cos ^{2} 75^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2} \theta+\sin ^{2} \theta}$
= 1 + 1
= 2

Question 45 

Find the value off the following:

$\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}$
Sol :
$\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}$

$=\sin \left\{90^{\circ}-\left(40^{\circ}+\theta\right)\right\}-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\sin ^{2}\left(90^{\circ}-50^{\circ}\right)}{\sin ^{2} 40^{\circ}+\cos ^{2}\left(90^{\circ}-50^{\circ}\right)}$

[∵ Sin θ = cos (90° – θ) and cos θ = sin (90° – θ)]

$=\sin \left(50^{\circ}-\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\sin ^{2}\left(40^{\circ}\right)}{\sin ^{2} 40^{\circ}+\cos ^{2}\left(40^{\circ}\right)}$
= 0 + 1
= 1

Question 46 

Find the value off the following:

$=\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 55^{\circ}, \cos \mathrm{ec} 35^{\circ}}{\tan 5^{\circ} \cdot \tan 25^{\circ} \cdot \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$
Sol :
$=\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 55^{\circ} \operatorname{cosec} 35^{\circ}}{\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$

$=\left(\frac{\sin \left(90^{\circ}-70^{\circ}\right)}{\sin 20^{\circ}}\right)+\frac{\sin \left(90^{\circ}-55^{\circ}\right) \operatorname{cosec} 35^{\circ}}{\cot \left(90^{\circ}-5^{\circ}\right) \cot \left(90^{\circ}-25^{\circ}\right) \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$
[∵ cos θ = sin (90° – θ) and tan θ = cot (90° – θ)]

$=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 35^{\circ} \operatorname{cosec} 35^{\circ}}{\cot 85^{\circ} \cot 65^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$

$=1+\frac{\sin 35^{\circ} \times \frac{1}{\sin 35^{\circ}}}{\frac{1}{\tan 85^{\circ}} \times \frac{1}{\tan 65^{\circ}} \times \tan 45^{\circ} \times \tan 65^{\circ} \times \tan 85^{\circ}}$
= 1 + 1 [∵ tan 45° = 1]
= 2

Question 47 

Find the value off the following:

$\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}+\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2}$
Sol :
$=\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}+\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2}$
$=\left(\frac{\sin 27^{\circ}}{\sin \left(90^{\circ}-63^{\circ}\right)}\right)^{2}+\left(\frac{\sin \left(90^{\circ}-63^{\circ}\right)}{\sin 27^{\circ}}\right)^{2}$ [∵ cos θ = sin (90° – θ)]
$=\left(\frac{\sin 27^{\circ}}{\sin 27^{\circ}}\right)^{2}+\left(\frac{\sin 27^{\circ}}{\sin 27^{\circ}}\right)^{2}$
= 1 + 1
= 2

Question 48 A 

Evaluate the following

$\frac{3 \sin 5^{\circ}}{\cos 85^{\circ}}+\frac{2 \cos 33^{\circ}}{\sin 57^{\circ}}$
Sol :
$=\frac{3 \sin 5^{\circ}}{\cos 85^{\circ}}+\frac{2 \cos 33^{\circ}}{\sin 57^{\circ}}$
$=\frac{3 \sin 5^{\circ}}{\sin \left(90^{\circ}-85^{\circ}\right)}+\frac{2 \sin \left(90^{\circ}-33^{\circ}\right)}{\sin 57^{\circ}}$ [∵ cos θ = sin (90° – θ)]
$=\frac{3 \sin 5^{\circ}}{\sin 5^{\circ}}+\frac{2 \sin 57^{\circ}}{\sin 57^{\circ}}$
= 3 + 2
= 5

Question 48 B 

Evaluate the following

$\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$
Sol :
$\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$
$=\frac{\cot 54^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}+\frac{\cot \left(90^{\circ}-20^{\circ}\right)}{\cot 70^{\circ}}-2$ [∵ tan θ = cot (90° – θ)]
$=\frac{\cot 54^{\circ}}{\cot 54^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}}-2$
= 1 +1 – 2
= 0

Question 48 C 

Evaluate the following

$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$
Sol :
$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$

$=\frac{\cos 80^{\circ}}{\cos \left(90^{\circ}-10^{\circ}\right)}+\sin \left(90^{\circ}-59^{\circ}\right) \operatorname{cosec} 31^{\circ}$
[∵ cos θ = sin (90° – θ) and sec θ = cosec (90° – θ)]

$=\frac{\cos 80^{\circ}}{\cos 80^{\circ}}+\sin 31^{\circ} \operatorname{cosec} 31^{\circ}$

$=1+\sin 31^{\circ} \times \frac{1}{\sin 31^{\circ}}$ $\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}$

= 1 + 1
= 2

Question 48 D 

Evaluate the following

cos38° cos52° – sin38° sin 52°
Sol :
We know that
cos θ = sin (90° – θ)
=sin (90° – 38°) sin (90° -52°) – sin 38° sin 52°
= sin 52° sin 38°– sin 38° sin 52°
=0

Question 48 E 

Evaluate the following

sec41° sin49° + cos49° cosec 41°
Sol :
We know that
sec θ = cosec (90° – θ) and cos θ = sin (90° – θ)
Cosec (90° – 41°) sin 49° + sin (90° – 49°) cosec 41°
= cosec 49° sin 49° + sin 41° cosec 41°

$=\frac{1}{\sin 49^{\circ}} \times \sin 49^{\circ}+\sin 41^{\circ} \times \frac{1}{\sin 41^{\circ}}$ $\left[\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}\right]$
= 1+ 1
= 2