ML Aggarwal Solutions for Class 9 Maths Chapter 17- Trigonometric Ratios

Class 9: Maths Chapter 17 solutions. Complete Class 9 Maths Chapter 17 Notes.

ML Aggarwal Solutions for Class 9 Maths Chapter 17- Trigonometric Ratios

ML Aggarwal 9th Maths Chapter 17, Class 9 Maths Chapter 17 solutions

Exercise 17

1. (a) From the figure (1) given below, find the values of:

(i) sin θ

(ii) cos θ

(iii) tan θ

(iv) cot θ

(v) sec θ

(vi) cosec θ

(b) From the figure (2) given below, find the values of:

(i) sin A

(ii) cos A

(iii) sin² A + cos² A

(iv) sec² A – tan² A.

Solution:

(a) From right angled triangle OMP,

By Pythagoras theorem, we get

OP² = OM² +MP²

MP² = OP² + OM²

MP² = (15)² – (12)²

MP² = 225 – 144

MP² = 81

MP² = 9²

MP = 9

(i) sin θ = MP/OP

= 9/15

= 3/5

(ii) cos θ = OM/OP

= 12/15

= 4/5

(iii) tan θ = MP/OP

= 9/12

= ¾

(iv) cot θ = OM/MP

= 12/9

= 4/3

(v) sec θ = OP/OM

= 15/12

= 5/4

(vi) cosec θ = OP/MP

= 15/9

= /3

(b) From right angled triangle ABC,

By Pythagoras theorem, we get

AB² = AC² + BC²

AB² = (12)² + (5)²

AB² = 144 + 25

AB² = 169

AB² = 13²

AB = 13

(i) sin A = BC/AB

= 5/13

(ii) cos A = AC/AB

= 12/13

(iii)Sin² A + cos² A = (BC/AB)² + (AC/AB)²

= (5/13)² + (12/13)²

= (25/169) + (144/169)

= (25 + 144)/ 169

= 169/169

= 1

Sin² A + cos² A = 1

(iv) Sec² A – tan² A = (AB/AC)² – (BC/AC)²

= (13/12)² – (5/12)²

= (169/144) – (25/144)

= (169 – 25)/ 144

= 144/144

= 1

Sec² A – tan² A = 1

2. (a) From the figure (1) given below, find the values of:
(i) sin B

(i) cos C

(iii) sin B + sin C

(iv) sin B cos C + sin C cos B.

(b) From the figure (2) given below, find the values of:

(i) tan x

(ii) cos y

(iii) cosec² y – cot² y

(iv) 5/sin x + 3/sin y – 3 cot y.

Solution:

From right angled triangle ABC,

By Pythagoras theorem, we get

BC² = AC² + AB²

AC² = BC² – AB²

AC² = (10)² – (6)²

AC² = 100 – 36

AC² = 64

AC² = 8²

AC = 8

(i) sin B = perpendicular/ hypotenuse

= AC/BC

= 8/10

= 4/5

(ii) cos C = Base/hypotenuse

= AC/BC

= 8/10

= 4/5

(iii) sin B = Perpendicular/hypotenuse

= AC/BC

= 8/10

= 4/5

Sin C = perpendicular/hypotenuse

= AB/BC

= 6/10

= 3/5

Now,

Sin B + sin C = (4/5) + (3/5)

= (4 + 3)/5

= 7/5

(iv) sin B = 4/5

Cos C = 4/5

Sin C = perpendicular/ hypotenuse

= AB/BC

= 6/10

= 3/5

Cos B = Base/Hypotenuse

= AB/BC

= 6/10

= 3/5

sin B cos C + sin C cos B

= (4/5) × (4/5) × (3/5) × (3/5)

= (26/25) × (9/25)

= (16+9)/25

= 25/25

= 1

From Figure

AC = 13, CD = 5, BC =21,

BD = BC – CD

= 21 – 5

= 16

From right angled ∆ACD,

By Pythagoras theorem we get

AC = AD² + CD²

AD² = AC² – CD²

AD² = (13)² – (5)²

AD² = 169 – 25

AD² = 144)

AD² = (12)²

AD = 12

From right angled ∆ABD,

By Pythagoras angled ∆ABD

By Pythagoras theorem we get

AB² = AD² + BD²

AB² = 400

AB² = (20)²

AB = 20

(i) tan x = perpendicular/Base (in right angled ∆ACD)

=CD/AD

= 5/12

(ii) cos y = Base/Hypotenuse (in right angled ∆ABD)

= BD/AB

= (20)/12 – (5/3)

Cot y = Base/Perpendicular (in right ∆ABD)

=BD/AB

= 16/20 = 4/5

(iii) cos y = Hypotenuse/ perpendicular (in right angled ∆ABD)

BD/AB

= 20/12

= 5/3

Cot y = Base/Perpendicular (in right ∆ABD)

AB/AD

= 16/12

= 4/3

Cosec² y – cot² y = (5/3)² – (4/3)²

= (25/9) – (16/9)

= (25-16)/9

= 9/9

= 1

Hence, cosec² y – cot² y = 1

(iv) sin x = Perpendicular/Hypotenuse (in right angled ∆ACD)

= AD/AB

= 12/20

= 3/5

Cot y = Base/Perpendicular (in right angled ∆ABD)

= BD/AD

= 16/12

= 4/3

(5/sin x) + (3/sin y) – 3cot y

= 5/(5/13) + 3/(3/5) – 3 × 4/3

= 5 × 13/5 + 3 × 5/3 – 3 × 4/3

= 1 × 13/1 + 1 × 5/1 – 1 × 4/1

= 13 + 5 – 4 = 18 – 4

= 14

Hence 5/sin x + 3/sin y – 3cot y = 14

3. (a) From the figure (1) given below, find the value of sec θ.

(b) From the figure (2) given below, find the values of:
(i) sin x

(ii) cot x

(iii) cot² x- cosec² x

(iv) sec y

(v) tan² y – 1/cos² y.

Solution:

(a) From the figure, Sec θ = AB / BD

But in ∆ADC, ∠D = 90^o

AC² =AD² + DC² (Pythagoras Theorem)

(13)² =AD² + 25

AD² = 169 -25

= 144

= (12)²

AD = 12

(in right ∆ABD)

AB² = AD² + BD²

= (12)² + (16)²

= 144 + 256

= 400

= (20)²

AB = 20

Now, Sec θ = AB / BD

= 20/16

= 5/4

(b) let given ∆ABC

BD = 3, AC = 12, AD = 4

In right angled ∆ABD

By Pythagoras theorem

AB² =AD² + BD²

AB² = (4)² + (3)²

AB² = 16 + 9

AB² = 25

AB² = (5)²

AB = 5

In right angled triangle ACD

By Pythagoras theorem,

AC² = AD² + CD²

CD² = AC² – AD²

CD² = (12)² – (4)²

CD² = 144 – 16

CD² = 128

CD = √128

CD = √64 × 2 CD

= 8√2

(i) sin x = perpendicular/Hypotenuse

= AD/AB

= 4/5

(ii) cot x = Base/Perpendicular

= BD/AD

= ¾

(iii) cot x = Base/ Perpendicular

BD/AD

= 3/4

(iv) cosec x = Hypotenuse / Perpendicular

AB/BD

= 5/4

Cot² x – cosec² x

= (3/4)² – (5/4)²

= 9/16 – 25/16

(9 -25)/16

= -16/16

= -1

Perpendicular = Hypotenuse/Base (in right angled ∆ACD)

= AD/CD

= 12/(8 √2)

= 3/(2 √2)

Cot y = Base/ Hypotenuse

= AD/CD

= 4/8 √ 2

= 1/2 √2

Cot y = Base / Hypotenuse (in right angled ∆ACD)

= CD/AC

= 8√2/12

= 2√/3

Now tan² y = 1/cos² y

= (1/2√2)² – 1/(2√2/3)²

= ¼ × – ¼ × 2

= (1/8) – (9/8)

= (1-9)/8

= -8/8

= -1

tan² y – 1/cos² y = – 1.

4. (a) From the figure (1) given below, find the values of:

(i) 2 sin y – cos y

(ii) 2 sin x – cos x

(iii) 1 – sin x + cos y

(iv) 2 cos x – 3 sin y + 4 tan x

(b) In the figure (2) given below, ∆ABC is right-angled at B. If AB = y units, BC = 3 units and CA = 5 units, find

(i) sin x⁰

(ii) y.

Solution:

(a) In a right angled ∆BCD,

Using Pythagoras theorem

BC² = BD² + CD²

Substituting the values

BC² = 9² + 12²

By further calculation

BC² = 81 + 144 = 225

BC² = 15²

BC = 15

In a right angled ∆ABC,

Using Pythagoras theorem

AC² = AB² + BC²

We can write it as

AB² = AC² – BC²

Substituting the values

AB² = 25² – 15²

By further calculation

AB² = 625 – 225 = 400

So we get

AB² = 20²

AB = 20

(i) We know that

In right angled ∆BCD

sin y = perpendicular/ hypotenuse

sin y = BD/ BC

Substituting the values

sin y = 9/15 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

2sin y – cos y = 2 × 3/5 – 4/5

We can write it as

= 6/5 – 4/5

= 2/5

Therefore, 2 sin y – cos y = 2/5

(ii) In right angled ∆ABC

sin x = perpendicular/ hypotenuse

sin x = BC/AC

Substituting the values

sin x = 15/25 = 3/5

In right angled ∆ABC

cos x = base/hypotenuse

cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

Here

2 sin x – cos x = 2 × 3/5 – 4/5

We can write it as

= 6/5 – 4/5

= 2/5

Therefore, 2 sin x – cos x = 2/5.

(iii) In right angled ∆ABC

sin x = perpendicular/hypotenuse

sin x = BC/AC

Substituting the values

sin x = 12/25 = 3/5

In right angled ∆BCD

cos y = base/hypotenuse

cos y = CD/BC

Substituting the values

cos y = 12/15 = 4/5

Here

1 – sin x + cos y = 1 – 3/5 + 4/5

By further calculation

= (5 – 3 + 4)/ 5

So we get

= (9 – 3)/ 5

= 6/5

Therefore, 1 – sin x + cos y = 6/5.

(iv) In right angled ∆BCD

cos x = base/hypotenuse

cos x = AB/AC

Substituting the values

cos x = 20/25 = 4/5

In right angled ∆BCD

sin y = perpendicular/hypotenuse

sin y = BD/BC

Substituting the values

sin y = 9/15 = 3/5

In right angled ∆ABC

tan x = perpendicular/base

tan x = BC/AB

Substituting the values

tan x = 15/20 = ¾

Here

2 cos x – 3 sin y + 4 tan x = 2 × 4/5 – 3 × 3/5 + 4 × ¾

By further calculation

= 8/5 – 9/5 3/1

Taking LCM

= (8 – 9 + 15)/5

So we get

= (23 – 9)/ 5

= 14/5

(b) It is given that

AB = y units, BC = 3 units, CA = 5 units

(i) In right angled ∆ABC

sin x = perpendicular/hypotenuse

sin x = BC/AC

Substituting the values

sin x = 3/5

(ii) In right angled ∆ABC

Using Pythagoras theorem

AC² = BC² + AB²

We can write it as

AB² = AC² – BC²

Substituting the values

AB² = 5² – 3²

By further calculation

AB² = 25 – 9 = 16

So we get

AB² = 4²

AB = 4

y = 4 units

Therefore, y = 4 units.

5. In a right-angled triangle, it is given that angle A is an acute angle and that
tan A=5/12. Find the values of:
(i) cos A
(ii) cosec A- cot A.

Solution:

Here, ABC is right angled triangle

∠A is an acute angle and ∠C = 90^o

tan A = 5/12

BC/AC =5/12

Let BC = 5x and AC = 12x

From right angled ∆ABC

By Pythagoras theorem, we get

AB² = (5x)² + (12x)²

AB² = 25x² + 144x²

AB² = 169x²

(i) cos A = Base/ Hypotenuse

= AC / AB

= 12x/13x

=12/13

(ii) cosec A = Hypotenuse/perpendicular

= AC / BC

= 13x /5x

= 13/5

cosec A – cot A = 13/5 – 12/5

= (13-12)/5

= 1/5

6. (a) In ∆ABC, ∠A = 90⁰. If AB = 7 cm and BC – AC = 1 cm, find:

(i) sin C

(ii) tan B

(b) In ∆PQR, ∠Q = 90⁰. If PQ = 40 cm and PR + QR = 50 cm, find:

(i) sin P

(ii) cos P

(iii) tan R.

Solution:

(a) In right ∆ABC

∠A = 90⁰

AB = 7 cm

BC – AC = 1 cm

BC = 1 + AC

We know that

BC² = AB² + AC²

Substituting the value of BC

(1 + AC)² = AB² + AC²

1 + AC² + 2AC = 7² + AC²

By further calculation

1 + AC² + 2AC = 49 AC²

2AC = 49 – 1 – 48

So we get

AC = 48/2 = 24 cm

Here

BC = 1 + AC

Substituting the value

BC = 1 + 24 = 25 cm

(i) sin C = AB/BC = 7/25

(ii) tan B = AC/AB = 24/7

(b) In right ∆PQR

∠Q = 90⁰

PQ = 40 cm

PQ + QR = 50 cm

We can write it as

PQ = 50 – QR

Using Pythagoras theorem

PR² = PQ² + QR²

(50 – QR)² = (40)² + QR²

By further calculation

2500 + QR² – 100QR = 1600 + QR²

So we get

2500 – 1600 = 100QR

100QR = 900

By division

QR = 900/100 = 9

We get

PR = 50 – 9 = 41

(i) sin P = QR/PR = 9/41

(ii) cos P = PQ/PR = 40/41

(iii) tan R = PQ/QR = 40/9

7. In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm. Find

(i) cos ∠ABC

(ii) sin ∠ACB.

Solution:

Here ABC is a triangle in which

AB = 15 cm, AC = 15 cm and BC = 18 cm

Draw AD perpendicular to BC , D is mid-point of BC.

Then, BD – DC = 9 cm

in right angled triangle ABD,

By Pythagoras theorem, we get

AB² = AD² + BD²

AD² = AB² – BD²

AD² = (15)² – (9)²

AD² = 225 – 81

AD² = 144

AD – 12 cm

(i) cos ∠ABC = Base/ / Hypotenuse

(In right angled ∆ABD, ∠ABC = ∠ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin ∠ACB = sin ∠ACD

= perpendicular/ Hypotenuse

= AD/AC

= 12/15

= 4/5

8. (a) In the figure (1) given below, ∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm. Find

(i) sin C

(ii) tan B

(iii) tan C – cot B.

(b) In the figure (2) given below, ∆ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin² θ + tan² θ.

(c) In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = 4/5 and tan C = 1.

(i) Calculate the lengths of AD, AB, DC and AC.

(ii) Show that tan² B – 1/cos² B = – 1.

Solution:

(a) It is given that

∆ABC is isosceles with AB = AC = 5 cm and BC = 6 cm

Construct AD perpendicular to BC

D is the mid point of BC

So BD = CD

Here

BD = CD = 6/2 = 3 cm

In right angled ∆ABD

Using Pythagoras theorem

AB² = AD² + BD²

We can write it as

AD² = AB² – BD²

Substituting the values

AD² = 5² – 3²

By further calculation

AD² = 25 – 9 = 16

So we get

AD² = 4²

AD = 4 cm

(i) In right angled ∆ACD

sin C = perpendicular/hypotenuse

sin C = AD/AC = 4/5

(ii) In right angled ∆ABD

tan B = perpendicular/base

tan B = AD/BD = 4/3

(iii) In right angled ∆ACD

tan C = perpendicular/base

tan C = AD/CD = 4/3

In right angled ∆ABD

cot B = base/perpendicular

cot B = BD/AD = ¾

Here

tan C – cot B = 4/3 – ¾

Taking LCM

tan C – cot B = (16 – 9)/ 12 = 7/12

(b) It is given that

∆ABC is right-angled at B

AB = 2 units and BC = 1 unit

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = 2² + 1²

AC² = 4 + 1 = 5

So we get

AC² = 5

AC = √5 units

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = 2/√5

In right angled ∆ABC

tan θ = perpendicular/base

tan θ = AB/BC = 2/1

We know that

sin² θ + tan² θ = (2/√5)² + (2/1)²

By further calculation

= 4/5 + 4/1

Taking LCM

= (4 + 20)/ 5

= 24/5

= 4 4/5

(c) (i) In ∆ABC

AD is perpendicular to BC

BD = 15 cm

sin B = 4/5

tan C = 1

In ∆ABD

sin B = perpendicular/hypotenuse

sin B = AD/AB = 4/5

Consider AD = 4x and AB = 5x

Using Pythagoras theorem

In right angled ∆ABD

AB² = AD² + BD²

We can write it as

BD² = AB² – AD²

Substituting the values

(15)² = (5x)² – (4x)²

225 = 25x² – 16x²

By further calculation

225 = 9x²

x² = 225/9 = 25

So we get

x = √25 = 5

Here

AD = 4 × 5 = 20

AB = 5 × 5 = 25

In right angled ∆ACD

tan C = perpendicular/base

So we get

tan C = AD/CD = 1/1

Consider AD = X then CD = x

In right angled ∆ADC

Using Pythagoras theorem

AC² = AD² + CD²

Substituting the values

AC² = x² + x² …..(1)

So the equation becomes

AC² = 20² + 20²

AC² = 400 + 400 = 800

So we get

AC = √800 = 20√2

Length of AD = 20 cm

Length of AB = 25 cm

Length of DC = 20 cm

Length of AC = 20√2 cm

(ii) In right angled ∆ABD

tan B = perpendicular/base

So we get

tan B = AD/BD

Substituting the values

tan B = 20/15 = 4/3

In right angled ∆ABD

cos B = base/hypotenuse

So we get

cos B = BD/AB

Substituting the values

cos B = 15/25 = 3/5

Here

LHS = tan² B – 1/cos² B

Substituting the values

= (4/3)² – 1/(3/5)²

By further calculation

= (4)²/(3)² – (5)²/(3)²

= 16/9 – 25/9

So we get

= (16 – 25)/9

= -9/9

= – 1

= RHS

Hence, proved.

9. If sin θ =3/5 and θ is acute angle, find

(i) cos θ

(ii) tan θ.

Solution:

Let ∆ ABC be a right angled at B

Let ∠ACB = θ

Given that, sin θ = 3/5

AB/AC = 3/5

Let AB = 3x

then AC = 5x

In right angled ∆ ABC,

By Pythagoras theorem,

We get

(5x)² = (3x)² + BC²

BC² = (5x)² – (3x)²

BC² = (2x)²

BC = 4x

(i) cos θ = Base/ Hypotenuse

= BC / AC

= 4x /5x

= 4/5

(ii) tan θ = perpendicular/Base

= AB/BC

= 3x/4x

= ¾

10. Given that tan θ = 5/12 and θ is an acute angle, find sin θ and cos θ.

Solution:

Consider ∆ ABC be right angled at B and ∠ACB = θ

It is given that

tan θ = 5/12

AB/BC = 5/12

Consider AB = 5x and BC = 12x

In right angled ∆ ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (5x)² + (12x)²

By further calculation

AC² = 25x² + 144x² = 169x²

So we get

AC² = (13x)²

AC = 13x

In right angled ∆ ABC

sin θ = perpendicular/hypotenuse

So we get

sin θ = AB/AC = 5x/13x = 5/13

In right angled ∆ ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC

Substituting the values

cos θ = 12x/13x = 12/13

11. If sin θ = 6/10, find the value of cos θ + tan θ.

Solution:

Consider ∆ ABC be right angled at B and ∠ACB = θ

It is given that

sin θ = AB/AC

sin θ = 6/10

Take AB = 6x then AC = 10x

In right angled ∆ ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

(10x)² = (6x)² + BC²

By further calculation

BC² = 100x² – 36x² = 64x²

So we get

BC² = (8x)²

BC = 8x

In right angled ∆ ABC

cos θ = base/hypotenuse

cos θ = BC/AC

Substituting the values

cos θ = 8x/10x = 4/5

In right angled ∆ ABC

tan θ = perpendicular/base

tan θ = AB/BC

Substituting the values

tan θ = 6x/8x = ¾

Here

cos θ + tan θ = 4/5 + ¾

Taking LCM

= (16 + 15)/ 20

= 31/20

= 1 11/20

12. If tan = 4/3, find the value of sin θ + cos θ (both sin θ and cos θ are positive).

Solution:

Let ∆ABC be a right angled

∠ACB = θ

Given that, tan θ = 4/3

(AB/BC = 4/3)

Give that, tan θ = 4/3

(AB/BC = 4/3)

Let AB = 4x,

then BC = 3x

In right angled ∆ABC

By Pythagoras theorem, we get

AC² = AB² + BC²

AC² = AB² + BC²

AC² = AB² + BC²

AC² = AB² + BC²

(AC² = (4x)² + (3x)²

AC² = 16x² + 9x²

AC² = 25x²

AC² = (5x)²

AC = 5x

Sin θ = perpendicular/Hypotenuse

= AB/AC

= 4x/5x

= 4/5

Cos θ = Base/Hypotenuse

= BC/AC

= 3x/5x

= 3/5

Sin θ + cos θ

= 4/5 + 3/5

= (4 + 3)/5

= 7/5

Hence, Sin θ + cos θ = 7/5 = 1 2/5

13. 1f cosec = √5 and θ is less than 90⁰, find the value of cot θ – cos θ.

Solution:

Given cosec θ = √5/1 = OP/PM

OP = √5 and PM = 1

Now OP² = OM² + PM² using Pythagoras theorem

(√5)² = OM² + 1²

5 = OM² + 1

OM² = 5 – 1

OM² = 4

OM = 2

Now cot θ = OM/PM

= 2/1

= 2

Cos θ = OM/OP

= 2/√5

Now cot θ – Cos θ = 2 – (2/√5)

= 2 (√5 – 1)/ √5

14. Given sin θ = p/q, find cos θ + sin θ in terms of p and q.

Solution:

Given that sin θ = p/q

Which implies,

AB/AC = p/q

Let AB = px

And then AC = qx

In right angled triangle ABC

By Pythagoras theorem,

We get

AC² = AB² + BC²

BC² = AC² – AB²

BC² = q²x² – p²x²

BC² = (q² – p²)x²

BC = √( q² – p²)x

In right angled triangle ABC,

Cos θ = base/ hypotenuse

= BC/AC

= √( q² – p²)x/qx

= √( q² – p²)/ q

Now,

Sin θ + cos θ = p/q + √( q² – p²)/ q

= [p + √( q² – p²)]/ q

15. If θ is an acute angle and tan = 8/15, find the value of sec θ + cosec θ.

Solution:

Given tan θ = 8/15

θ is an acute angle

in the figure triangle OMP is a right angled triangle,

∠M = 90^o and ∠Q = θ

Tanθ = PM/OL = 8/15

Therefore, PM = 8, OM = 15

But OP² = OM² + PM² using Pythagoras theorem,

= 15² + 8²

= 225 + 64

= 289

= 17²

Therefore, OP = 17

Sec θ = OP/OM

= 17/15

Cosec θ = OP/PM

= 17/8

Now,

Sec θ + cosec θ = (17/15) + (17/8)

= (136 + 255)/ 120

= 391/120

= 3 31/120

16. Given A is an acute angle and 13 sin A = 5, evaluate:

(5 sin A – 2 cos A)/ tan A.

Solution:

Let triangle ABC be a right angled triangle at B and A is an acute angle

Given that 13 sin A = 5

Sin A = 5/13

AB/Ac = 5/13

Let AB = 5x

AC = 13 x

In right angled triangle ABC,

Using Pythagoras theorem,

We get

AC² = AB² + BC²

BC² = AC² – BC²

BC² = (13x)² – (5x)²

BC² = 169x² – 25x²

BC² = 144x²

BC = 12x

Sin A = 5/13

Cos A = base/ hypotenuse

= BC/AC

= 12x/ 13x

= 12/13

Tan A = perpendicular/ base

= AB/BC

= 5x/ 12x

= 5/ 12

Now,

(5 sin A – 2 cos A)/ tan A = [(5) (5/13) – (2) (12/13)]/ (5/12)

= (1/13)/ (5/12)

= 12/65

Hence (5 sin A – 2 cos A)/ tan A = 12/65

17. Given A is an acute angle and cosec A = √2, find the value of

(2 sin² A + 3 cot² A)/ (tan² A – cos² A).

Solution:

Let triangle ABC be a right angled at B and A is a acute angle.

Given that cosec A = √2

Which implies,

AC/BC = √2/1

Let AC = √2x

Then BC = x

In right angled triangle ABC

By using Pythagoras theorem,

We get

AC² = AB² + BC²

(√2x)² = AB² + x²

AB² = 2x² – x²

AB = x

Sin A = perpendicular/ hypotenuse

= BC/AC

= 1/ √2

Cot A = base/ perpendicular

= x/x

= 1

Tan A = perpendicular/ base

= BC/AB

= x/x

= 1

Cos A = base/ hypotenuse

= AB/AC

= x/ √2x

= 1/√2

Substituting these values we get

2 sin² A + 3 cot² A/ (tan² A – cos² A) = 8

18. The diagonals AC and BD of a rhombus ABCD meet at O. If AC = 8 cm and BD = 6 cm, find sin ∠OCD.

Solution:

It is given that

Diagonals AC and BD of rhombus ABCD meet at O

AC = 8 cm and BD = 6 cm

O is the mid point of AC

We know that

AO = OC = AC/2 = 8/2 = 4 cm

O is the mid point of BD

BO = OD = BD/2 = 6/2 = 3 cm

In right angled ∆COD

CD² = OC² + OD²

Substituting the values

CD² = 4² + 3²

So we get

CD² = 16 + 9 = 25

CD² = 5²

CD = 5 cm

In right angled ∆COD

sin ∠OCD = perpendicular/ hypotenuse

So we get

sin ∠OCD = OD/CD = 3/5

19. If tan θ = 5/12, find the value of (cos θ + sin θ)/ (cos θ – sin θ).

Solution:

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

tan θ = AB/BC = 5/12

Take AB = 5x then BC = 12x

In right angled ∆ABC,

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (5x)² + (12x)²

By further calculation

AC² = 25x² + 144x² = 169x²

So we get

AC² = (13x)²

AC = 13x

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC

Substituting the values

cos θ = 12x/13x = 12/13

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC

Substituting the values

sin θ = 5x/13x = 5/13

Here

(cos θ + sin θ)/ (cos θ – sin θ) = [12/13 + 5/13]/ [12/13 – 5/13]

Taking LCM

= [(12 + 5)/ 13]/[(12 – 5)/ 13]

So we get

= 17/13/ 7/13

= 17/13 × 13/7

= 17/7

Therefore, (cos θ + sin θ)/ (cos θ – sin θ) = 17/7 = 2 3/7.

20. Given 5 cos A – 12 sin A = 0, find the value of (sin A + cos A)/ (2 cos A – sin A).

Solution:

It is given that

5 cos A – 12 sin A = 0

We can write it as

5 cos A = 12 sin A

So we get

sin A/ cos A = 5/12

We know that sin A/ cos A = tan A

tan A = 5/12

Consider ∆ABC right angled at B and ∠A is acute angle

Here

tan A = BC/AB = 5/12

Take BC = 5x then AB = 12x

In right angled ∆ABC

Using Pythagoras theorem

AC² = BC² + AB²

Substituting the values

AC² = (5x)² + (12x)²

AC² = 25x² + 144x² = 169x²

So we get

AC² = (13x)²

AC = 13x

In right angled ∆ABC

sin A = perpendicular/hypotenuse

So we get

sin A = BC/AC = 5x/13x = 5/13

In right angled ∆ABC

cos A = base/hypotenuse

So we get

cos A = AB/AC = 12x/13x = 12/13

Here

(sin A + cos A)/ (2 cos A – sin A) = [5/13 + 12/13]/ [2 × 12/13 – 5/13]

By further calculation

= [(5 + 12)/13]/ [24/13 – 5/13]

So we get

= [(5 + 12)/13]/[(24 – 5)/13]

= 17/13/ 19/13

= 17/13 × 13/19

= 17/19

Therefore, (sin A + cos A)/ (2 cos A – sin A) = 17/19

21. If tan θ = p/q, find the value of (p sin θ – q cos θ)/ (p sin θ + q cos θ).

Solution:

It is given that

tan θ = p/q

Consider ∆ABC be right angled at B and ∠BCA = θ

tan θ = BC/AB = p/q

BC = px then AB = qx

In right angled ∆ABC

Using Pythagoras theorem

AC² = BC² + AB²

Substituting the values

AC² = (px)² + (qx)²

AC² = p²x² + q²x²

AC² = x² (p² + q²)

So we get

AC = √x² (p² + q²)

AC = x(√p² + q²)

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = BC/AC

Substituting the values

sin θ = px/ x(√p² + q²)

So we get

sin θ = p/ (√p² + q²)

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = AB/AC

Substituting the values

cos θ = qx/ x(√p² + q²)

So we get

cos θ = q/ (√p² + q²)

Here

22. If 3 cot θ = 4, find the value of (5 sin θ – 3 cos θ)/ (5 sin θ + 3 cos θ).

Solution:

It is given that

3 cot θ = 4

cot θ = 4/3

Consider ∆ABC be right angled at B and ∠ACB = θ

cot θ = BC/AB = 4/3

Take BC = 4x then AB = 3x

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (3x)² + (4x)²

AC² = 9x² + 16x² = 25x²

So we get

AC² = (5x)²

AC = 5x

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC

Substituting the values

sin θ = 3x/5x = 3/5

In right angled ∆ABC

cos θ = base/hypotenuse

cos θ = BC/AC

Substituting the values

cos θ = 4x/5x = 4/5

23. (i) If 5 cos θ – 12 sin θ = 0, find the value of (sin θ + cos θ)/(2 cos θ – sin θ).

(ii) If cosec θ = 13/12, find the value of (2 sin θ – 3 cos θ)/(4 sin θ – 9 cos θ).

Solution:

(i) It is given that

5 cos θ – 12 sin θ = 0

We can write it as

5 cos θ = 12 sin θ

sin θ/ cos θ = 5/12

tan θ = 5/12

Dividing both numerator and denominator by cos θ

(ii) It is given that

cosec θ = 13/12

We know that cosec θ = 1/ sin θ

1/sin θ = 13/12

sin θ = 12/13

Here cos² θ = 1 – sin² θ

Substituting the values

= 1 – (12/13)²

By further calculation

= 1 – 144/169

Taking LCM

= (169 – 144)/ 169

= 25/169

So we get

= (5/13)²

cos θ = 5/13

Here

24. If 5 sin θ = 3, find the value of (sec θ – tan θ)/ (sec θ + tan θ).

Solution:

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

5 sin θ = 3

sin θ = AB/AC = 3/5

Take AB = 3x then AC = 5x

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

BC² = AC² – AB²

Substituting the values

BC² = (5x)² – (3x)²

So we get

BC² = 25x² – 9x² = 16x²

BC² = (4x)²

BC = 4x

In right angled ∆ABC

sec θ = hypotenuse/base

sec θ = AC/BC = 5x/4x = 5/4

In right angled ∆ABC

tan θ = perpendicular/base

tan θ = AB/BC = 3x/4x = ¾

25. If θ is an acute angle and sin θ = cos θ, find the value of 2 tan² θ + sin² θ – 1.

Solution:

Consider ∆ABC be right angled at B and ∠ACB = θ

It is given that

sin θ = cos θ

sin θ/cos θ = 1

tan θ = AB/BC = 1

Take AB = x then BC = x

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

AC² = x² + x² = 2x²

So we get

AC = √2x²

AC = (√2)x

In right angled ∆ABC

sin θ = perpendicular/hypotenuse

sin θ = AB/AC = x/√2x = 1/√2

Here

2 tan² θ + sin² θ – 1 = 2 × (1)² + (1/√2)² – 1

By further calculation

= 2 × 1 + ½ – 1

= 2 + ½ – 1

= 1+ ½

Taking LCM

= (2 + 1)/2

= 3/2

Therefore, 2 tan² θ + sin² θ – 1 = 3/2.

26. Prove the following:

(i) cos θ tan θ = sin θ

(ii) sin θ cot θ = cos θ

(iii) sin² θ/ cos θ + cos θ = 1/ cos θ.

Solution:

(i) cos θ tan θ = sin θ

LHS = cos θ tan θ

We know that tan θ = sin θ/cos θ

= cos θ (sin θ/cos θ)

So we get

= 1 × sin θ/1

= sin θ

= RHS

Therefore, LHS = RHS.

(ii) sin θ cot θ = cos θ

LHS = sin θ cot θ

We know that cot θ = cos θ/sin θ

= sin θ (cos θ/sin θ)

= 1 × cos θ/1

= cos θ

= RHS

Therefore, LHS = RHS.

(iii) sin² θ/cos θ + cos θ = 1/cos θ

LHS = sin² θ/cos θ + cos θ/1

Taking LCM

= (sin² θ + cos² θ)/cos θ

We know that sin² θ + cos² θ = 1

= 1/cos θ

= RHS

Therefore, LHS = RHS.

27. If in ∆ABC, ∠C = 90⁰ and tan A = ¾, prove that sin A cos B + cos A sin B = 1.

Solution:

It is given that

tan A = BC/AC = ¾

Using Pythagoras theorem

AB² = AC² + BC²

Substituting the values

= 4² + 3²

= 16 + 9

= 25

= 5²

So we get AB = 5

Here

sin A = BC/AC = 3/5

cos A = AC/AB = 4/5

cos B = BC/AB = 3/5

sin B = AC/AB = 4/5

LHS = sin A cos B + cos A sin B

Substituting the values

= 3/5 × 3/5 + 4/5 × 4/5

By further calculation

= 9/25+ 16/25

= (9 + 16)/ 25

= 25/25

= 1

= RHS

Therefore, LHS = RHS.

28. (a) In figure (1) given below, ∆ABC is right-angled at B and ∆BRS is right-angled at R. If AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x⁰ and ∠SAB = y⁰, then find:

(i) tan x⁰

(ii) sin y⁰.

(b) In the figure (2) given below, ∆ABC is right angled at B and BD is perpendicular to AC. Find

(i) cos ∠CBD

(ii) cot ∠ABD.

Solution:

(a) ∆ABC is right-angled at B, ∆BSC is right-angled at S and ∆BRS is right-angled at R

It is given that

AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x⁰ and ∠SAB = y⁰

By Geometry ∆ARS and ∆ABC are similar

AR/AB = RS/BC

Substituting the values

AR/18 = 5/7.5

By further calculation

AR = (5 × 18)/7.5 = (1 × 18)/1.5

Multiply both numerator and denominator by 10

AR = (18 × 10)/15

AR = (10 × 6)/5

AR = (2 × 6)/1 = 12

So we get

RB = AB – AR

RB = 18 – 12 = 6

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = 18² + 7.5²

By further calculation

AC² = 324 + 56.25 = 380.25

AC = √380.25 = 19.5 cm

(i) In right angled ∆BSR

tan x⁰ = perpendicular/base

tan x⁰ = RB/RS = 6/5

(ii) In right angled ∆ASR

sin y⁰ = perpendicular/hypotenuse

Using Pythagoras theorem

AS² = 12² + 5²

By further calculation

AS² = 144 + 25 = 169

AS = √169 = 13 cm

So we get

sin y⁰ = RS/AS = 5/13

(b) We know that

∆ABC is right angled at B and BD is perpendicular to AC

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = 12² + 5²

By further calculation

AC² = 144 + 25 = 169

So we get

AC² = (13)²

AC = 13

By Geometry ∠CBD = ∠A and ∠ABD = ∠C

(i) cos ∠CBD = cos ∠A = base/hypotenuse

In right angled ∆ABC

cos ∠CBD = cos ∠A = AB/AC = 12/13

(ii) cos ∠ABD = cos ∠C = base/perpendicular

In right angled ∆ABC

cos ∠ABD = cos ∠C = BC/AB = 5/12

29. In the adjoining figure, ABCD is a rectangle. Its diagonal AC = 15 cm and ∠ACD = α. If cot α = 3/2, find the perimeter and the area of the rectangle.

Solution:

In right ∆ADC

cot α = CD/AD = 3/2

Take CD = 3x then AD = 2x

Using Pythagoras theorem

AC² = CD² + AD²

Substituting the values

(15)² = (3x)² + (2x)²

By further calculation

13x² = 225

x² = 225/13

So we get

x = √225/13 = 15/√13

Length of rectangle (l) = 3x = (3 × 15)/ √13 = 45/√13 cm

Breadth of rectangle (b) = 2x = (2 × 15)/ √13 = 30/√13 cm

(i) Perimeter of rectangle = 2 (l + b)

Substituting the values of l and b

= 2 (45/√13 + 30/√13)

So we get

= 2 × 75/√13

= 150/√13 cm

(ii) Area of rectangle = l × b

Substituting the values of l and b

= 45/√13 × 30/√13

So we get

= 1350/13

= 103 11/13 cm²

30. Using the measurements given in the figure alongside,

(a) Find the values of:

(i) sin ϕ

(ii) tan θ.

(b) Write an expression for AD in terms of θ.

Solution:

From the figure

BC = 12, BD = 13

In right angled ∆BCD

Using Pythagoras theorem

BD² = BC² + CD²

It can be written as

CD² = BD² – BC²

Substituting the values

CD² = (13)² – (12)²

CD² = 169 – 144 = 25

So we get

CD = √25 = 5

Construct BE perpendicular to AB

CD = BE = 5 and EA = AE = 14 – 5 = 9

(a) (i) sin ϕ = perpendicular/hypotenuse

In right angled ∆BCD

sin ϕ = CD/BD = 5/13

(ii) tan θ = perpendicular/hypotenuse

In right angled ∆AED

tan θ = ED/AE = BC/AE = 12/9 = 4/3 (Since ED = BC)

(b) In right angled ∆AED

sin θ = perpendicular/hypotenuse

cos θ = base/perpendicular

We can write it as

sin θ = ED/AD or cos θ = AE/AD

AD = ED/sin θ or AD = AE/ cos θ

Substituting the values

AD = 12/sin θ or AD = 9/cos θ

Therefore, AD = 12/ sin θ or AD = 9/cos θ.

31. Prove the following:

(i) (sin A + cos A)² + (sin A – cos A)² = 2

(ii) cot² A – 1/sin² A + 1 = 0

(iii) 1/(1 + tan² A) + 1/(1 + cot² A) = 1

Solution:

(i) (sin A + cos A)² + (sin A – cos A)² = 2

LHS = (sin A + cos A)² + (sin A – cos A)²

Using the formula

(a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab

= [(sin A)² + (cos A)² + 2 sin A cos A] + [(sin A)² + (cos A)² – 2 sin A cos A]

By further calculation

= sin² A + cos² A + 2 sin A cos A + sin² A + cos² A – 2 sin A cos A

= sin² A + cos² A + sin² A + cos² A

= 2 sin² A + 2 cos² A

We know that sin² A + cos² A = 1

= 2 (sin² A + cos² A)

= 2 (1)

= 2

= RHS

Therefore, LHS = RHS.

(ii) cot² A – 1/sin² A + 1 = 0

LHS = cot² A – 1/sin² A + 1

We know that

1/sin A = cosec A

= cot² A – cosec² A + 1

= (1 + cot² A) – cosec² A

We know that 1 + cot² A = cosec² A

= cosec² A – cosec² A

= 0

= RHS

Therefore, LHS = RHS.

(iii) 1/(1 + tan² A) + 1/(1 + cot² A) = 1

LHS = 1/(1 + tan² A) + 1/(1 + cot² A)

We know that

sec² A – tan² A = 1

sec² A = 1 + tan² A

cosec² A – cot² A = 1

cosec² A = 1 + cot² A

So we get

= 1/sec² A + 1/cosec² A

Here 1/sec A = cos A and 1/cosec A = sin A

= cos² A + sin² A

= 1

= RHS

Therefore, LHS = RHS.

32. Simplify

Solution:

We know that 1 = sin² θ + cos² θ

= √cos² θ/sin² θ

= cos θ/ sin θ

Here cos θ/sin θ = cot θ

= cot θ

Therefore,
= cot θ.

33. If sin θ + cosec θ = 2, find the value of sin² θ + cosec² θ.

Solution:

It is given that

sin θ + cosec θ = 2

sin θ + 1/sin θ = 2

By further calculation

sin² θ + 1 = 2 sin θ

sin² θ – 2 sin θ + 1 = 0

So we get

(sin θ – 1)² = 0

sin θ – 1 = 0

sin θ = 1

Here

sin² θ + cosec² θ = sin² θ + 1/sin² θ

Substituting the values

= 1² + 1/1²

= 1 + 1/1

= 1 + 1

= 2

34. If x = a cos θ + b sin θ and y = a sin θ – b cos θ, prove that x² + y² = a² + b².

Solution:

It is given that

x = a cos θ + b sin θ …. (1)

y = a sin θ – b cos θ …. (2)

By squaring and adding both the equations

x² + y² = (a cos θ + b sin θ)² + (a sin θ – b cos θ)²

Using the formula

(a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab

= [(a cos θ)² + (b sin θ)² + 2 (a cos θ) (b sin θ)] + [(a sin θ)² + (b cos θ)² – 2 (a sin θ) (b cos θ)]

By further calculation

= a² cos² θ + b² sin² θ + 2 ab sin θ cos θ + a² sin² θ + b² cos² θ – 2 ab sin θ cos θ

= a² cos² θ + b² sin² θ + a² sin² θ + b² cos² θ

So we get

= a² (cos² θ + sin² θ) + b² (sin² θ + cos² θ)

Here sin² θ + cos² θ = 1

= a² (1) + b² (1)

= a² + b²

Therefore, x² + y² = a² + b².

Chapter test

1. (a)From the figure (i) given below, calculate all the six t-ratios for both acute………

(b)From the figure (ii) given below, find the values of x and y in terms of t-ratios

Solution:

(a) From right angled triangle ABC,

By Pythagoras theorem, we get

AC² = AB² + BC²

AB² = AC² – BC²

AB² = (3)² – (2)²

AB² = 9 – 4

AB² = 5

AB = √5

(i) sin A = perpendicular/ hypotenuse

= BC/AC

= 2/3

(ii) cos A = base/ hypotenuse

= AB/AC

= √5/3

(iii) tan A = perpendicular/ base

= BC/AB

= 2/ √5

(iv) cot A = base/perpendicular

= AB/ BC

= √5/2

(v) sec A = hypotenuse/ base

= AC/AB

= 3/ √5

(vi) cosec A = hypotenuse/perpendicular

= AC/BC

= 3/2

(b) From right angled triangle ABC,

∠BAC = θ

Then we know that,

Cot θ = base/ perpendicular

= AB/ BC

= x/ 10

x = 10 cot θ

also, cosec θ = hypotenuse/ perpendicular

= AC/ BC

= y/ 10

y = 10 cosec θ

Therefore, x = 10 cot θ and y = 10 cosec θ.

2. (a) From the figure (1) given below, find the values of:

(i) sin ∠ABC

(ii) tan x – cos x + 3 sin x.

(b) From the figure (2) given below, find the values of:

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x – 17 sin y – tan x.

Solution:

(a) From the figure

BC = 12, CD = 9 and BC = 20

In right angled ∆ABC,

Using Pythagoras theorem

AB² = AC² + BC²

It can be written as

AC² = AB² – BC²

Substituting the values

AC² = (20)² – (12)²

By further calculation

AC² = 400 – 144 = 256

So we get

AC² = (16)²

AC = 16

In right angled ∆BCD

Using Pythagoras theorem

BD² = BC² + CD²

Substituting the values

BD² = 12² + 9²

By further calculation

BD² = 144 + 81 = 225

So we get

BD² = (15)²

BD = 15

(i) In right angled ∆BCD

sin ∠ABC = perpendicular/hypotenuse

So we get

sin ∠ABC = AC/AB = 16/20 = 4/5

(ii) In right angled ∆BCD

tan x = perpendicular/base

So we get

tan x = BC/CD = 12/9 = 4/3

In right angled ∆BCD

cos x = base/hypotenuse

So we get

cos x = CD/BD = 9/15 = 3/5

In right angled ∆BCD

sin x = perpendicular/hypotenuse

So we get

sin x = BC/BD = 12/15 = 4/5

tan x – cos x + 3 sin x = 4/3 – 3/5 + 3 × 4/5

By further calculation

= 4/3 – 3/5 + 12/5

Taking LCM

= (4 × 5 – 3 × 3 + 12 × 3)/ 15

So we get

= (20 – 9 + 36)/ 15

= (56 – 9)/ 15

= 27/15

= 3 2/15

Therefore, tan x – cos x + 3 sin x = 3 2/15.

(b) In the figure

AC = 17, AB = 25, AD = 15

In right angled ∆ACD

Using Pythagoras theorem

AC² = AD² + CD²

Substituting the values

(17)² = (15)² + (CD)²

By further calculation

CD² = (17)² – (15)²

CD² = 289 – 225 = 64

So we get

CD² = 8²

CD = 8

In right angled ∆ABD

Using Pythagoras theorem

AB² = AD² + BD²

Substituting the values

(25)² = (15)² + BD²

By further calculation

BD² = (25)² – (15)²

BD² = 625 – 225 = 400

So we get

BD² = (20)²

BD = 20

(i) In right angled ∆ABD

5 sin x = 5 (perpendicular/hypotenuse)

So we get

= 5 (AD/AB)

= 5 × 15/25

= 15/5

= 3

(ii) In right angled ∆ABD

7 tan x = 7 (perpendicular/base)

So we get

= 7 (AD/AB)

= 7 × 15/20

= 7 × ¾

= 21/4

= 5 ¼

(iii) In right angled ∆ABD

cos x = base/hypotenuse

So we get

cos x = BD/AB = 20/25 = 4/5

In right angled ∆ACD

sin y = perpendicular/hypotenuse

So we get

sin y = CD/AC = 8/17

In right angled ∆ABD

tan x = perpendicular/base

So we get

tan x = AD/BD = 15/20 = ¾

5 cos x – 17 sin y – tan x = 5 × 4/5 – 17 × 8/17 – ¾

It can be written as

= 4/1 – 8/1 – ¾

Taking LCM

= (16 – 32 – 3)/4

= (16 – 35)/4

So we get

= – 19/4

= – 4 ¾

Therefore, 5 cos x – 17 sin y – tan x = – 4 ¾.

3. If q cos θ = p, find tan θ – cot θ in terms of p and q.

Solution:

Consider ABC as a triangle right angled at B and ∠ACB = θ

It is given that

q cos θ = p

cos θ = BC/AC = p/q

Take BC = px then AC = qx

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

It can be written as

AB² = AC² – BC²

Substituting the values

AB² = (qx)² – (px)²

AB² = q²x² – p²x²

Taking out the common terms

AB² = (q² – p²)x²

So we get

AB = √(q² – p²) x²

AB = (√q² – p²)x

In right angled ∆ABC

tan θ = perpendicular/base

So we get

tan θ = AB/BC = [(√q² – p²)x]/px

tan θ = (√q² – p²)/p

In right angled ∆ABC

cot θ = base/perpendicular

So we get

cot θ = BC/AB = px/[(√q² – p²)x]

cot [(√q² – p²)x] = p/(√q² – p²)

4. Given 4 sin θ = 3 cos θ, find the values of:

(i) sin θ

(ii) cos θ

(iii) cot² θ – cosec² θ.

Solution:

It is given that

4 sin θ = 3 cos θ

sin θ/cos θ = ¾

tan θ = ¾

Consider ∆ABC right angled at B and ∠ACB = θ

tan θ = perpendicular/base

Substituting the values

¾ = AB/BC

AB/BC = ¾

Take AB = 3x then BC = 4x

In right angled ∆ABC

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

AC² = (3x)² + (4x)²

By further calculation

AC² = 9x² + 16x² = 25x²

So we get

AC² = (5x)²

AC = 5x

(i) In right angled ∆ABC

sin θ = perpendicular/hypotenuse

So we get

sin θ = AB/AC = 3x/5x = 3/5

(ii) In right angled ∆ABC

cos θ = base/hypotenuse

So we get

cos θ = BC/AC = 4x/5x = 4/5

(iii) In right angled ∆ABC

cot θ = base/perpendicular

So we get

cot θ = BC/AB = 4x/3x = 4/3

In right angled ∆ABC

cosec θ = hypotenuse/perpendicular

So we get

cosec θ = AC/AB = 5x/3x = 5/3

Here

cot² θ – cosec² θ = (4/3)² – (5/3)²

By further calculation

= 16/9 – 25/9

= (16 – 25)/9

= -9/9

= – 1

Therefore, cot² θ – cosec² θ = -1.

5. If 2 cos θ = √3, prove that 3 sin θ – 4 sin³ θ = 1.

Solution:

It is given that

2 cos θ = √3

cos θ = √3/2

We know that

sin² θ = 1 – cos² θ

Substituting the values

= 1 – (√3/2)²

= 1 – ¾

= ¼

sin θ = √ ¼ = ½

Consider

LHS = 3 sin θ – 4 sin³ θ

It can be written as

= sin θ (3 – 4 sin² θ)

Substituting the values

= ½ (3 – 4 × ¼)

= ½ (3 – 1)

= ½ × 1

= 1

= RHS

Therefore, proved.

6. If (sec θ – tan θ)/ (sec θ + tan θ) = ¼, find sin θ.

Solution:

We know that

By cross multiplication

4 – 4 sin θ = 1 + sin θ

We get

4 – 1 = sin θ + 4 sin θ

3 = 5 sin θ

sin θ = 3/5

7. If sin θ + cosec θ = 3 1/3, find the value of sin² θ + cosec² θ.

Solution:

It is given that

sin θ + cosec θ = 3 1/3 = 10/3

By squaring on both sides

(sin θ + cosec θ)² = (10/3)²

Expanding using formula (a + b)² = a² + b² + 2ab

sin² θ + cosec² θ + 2 sin θ cosec θ = 100/9

We know that sin θ = 1/cosec θ

sin² θ + cosec² θ + 2 sin θ × 1/ sin θ = 100/9

By further calculation

sin² θ + cosec² θ + 2 = 100/9

sin² θ + cosec² θ = 100/9 – 2

Taking LCM

sin² θ + cosec² θ = (100 – 18)/9 = 82/9

So we get

sin² θ + cosec² θ = 9 1/9

8. In the adjoining figure, AB = 4 m and ED = 3 m.

If sin α = 3/5 and cos β = 12/13, find the length of BD.

Solution:

It is given that

sin α = AB/AC = 3/5

AB = 3 and AC = 5

Using Pythagoras theorem

AC² = AB² + BC²

Substituting the values

5² = 3² + BC²

By further calculation

25 = 9 + BC²

BC² = 25 – 9 = 16

So we get

BC² = 4²

BC = 4

We know that

tan α = AB/BC = 4/5

cos β = CD/CE = 12/13

CD = 12 and CE = 13

Using Pythagoras theorem

CE² = CD² + ED²

Substituting the values

13² = 12² + ED²

By further calculation

ED² = 13² – 12²

ED² = 169 – 144 = 25

So we get

ED² = (5)²

ED = 5

tan β = ED/CD = 5/12

From the figure

tan α = AB/BC = 4/BC

So we get

¾ = 4/BC

BC = (4 × 4)/3 = 16/3 m

tan β = ED/CD = 3/CD

5/12 = 3/CD

So we get

CD = (12 × 3)/5 = 36/5 m

Here

BD = BC + CD

Substituting the values

= 16/3 + 36/5

Taking LCM

= (80 + 108)/15

= 188/15 m

= 12 8/15 m

Download PDF

ML Aggarwal Solutions for Class 9 Maths Chapter 17- Trigonometric Ratios

Download PDF: ML Aggarwal Solutions for Class 9 Maths Chapter 17- Trigonometric Ratios PDF

Chapterwise ML Aggarwal Solutions for Class 9 Maths :

• Chapter 1- Rational and Irrational Numbers
• Chapter 2- Compound Interest
• Chapter 3- Expansions
• Chapter 4- Factorization
• Chapter 5- Simultaneous Linear Equations
• Chapter 6- Problems on Simultaneous Linear Equations
• Chapter 7- Quadratic Equations
• Chapter 8- Indices
• Chapter 9- Logarithms
• Chapter 10- Triangles
• Chapter 11- Mid Point Theorem
• Chapter 12- Pythagoras Theorem
• Chapter 13- Rectilinear Figures
• Chapter 14- Theorems on Area
• Chapter 15- Circle
• Chapter 16- Mensuration
• Chapter 17- Trigonometric Ratios
• Chapter 18- Trigonometric Ratios and Standard Angles
• Chapter 19- Coordinate Geometry
• Chapter 20- Statistics

About ML Aggarwal

M. L. Aggarwal, is an Indian mechanical engineer, educator. His achievements include research in solutions of industrial problems related to fatigue design. Recipient Best Paper award, Manipal Institute of Technology, 2004. Member of TSTE.