# Frank Solutions for Chapter 26 Trigonometric Ratios Class 9 Mathematics ICSE

### Exercise 26.1

1. In each of the following, one trigonometric ratio is given. Find the values of the other trigonometric ratios.

(i) sin A = 12/13

(ii) cos B = 4/5

(iii) cot A = 1/11

(iv) cosec C = 15/11

(v) tan C = 5/12

(vi) sin B = √3/2

(vii) cos A = 7/25

(viii) tan B = 8/15

(ix) sec B = 15/12

(x) cosec C = √10

(i) Sin A = 12/13

sin A = Perpendicular/Hypotenuse

⇒ sin A = 12/13

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Base = √(Hypotenuse)2 – (Perpendicular)2

⇒ Base = √(13)2 – (12)2

⇒ Base = √169 – 144

⇒ Base = √25

We get,

Base = 5

cos A = Base/Hypotenuse

⇒ cos A = 5/13

sec A = (1/cos A)

⇒ sec A = 13/5

cot A = (1/tan A)

⇒ cot A = 5/12

cosec A = (1/sin A)

⇒ cosec A = 13/12

(ii) cos B = 4/5

⇒ cos B = Base/Hypotenuse

⇒ cos B = 4/5

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Perpendicular = √(Hypotenuse)2 – (Base)2

⇒ Perpendicular = √(5)2 – (4)2

⇒ Perpendicular = √25 – 16

⇒ Perpendicular = √9

We get,

Perpendicular = 3

sin B = Perpendicular/Hypotenuse

⇒ sin B = 3/5

tan B = Perpendicular/Base

⇒ tan B = 3/4

sec B = (1/cos B)

⇒ sec B = 5/4

cot B = (1/tan B)

⇒ cot B = 4/3

cosec B = (1/sin B)

⇒ cosec B = 5/3

(iii) cot A = 1/11

cot A = (1/tan A)

cot A = Base/Perpendicular

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

(Hypotenuse) = √(Perpendicular)2 + (Base)2

⇒ (Hypotenuse) = √(11)2 + (1)2

⇒ (Hypotenuse) = √(121 + 1)

⇒ (Hypotenuse) = √122

cos A = Base/Hypotenuse

⇒ cos A = 1/√122

tan A = Perpendicular/Base

⇒ tan A = 11

sec A = (1/cos A)

⇒ sec A = √122

sin A = Perpendicular/Hypotenuse

⇒ sin A = 11/√122

cosec A = (1/sin A)

⇒ cosec A = √122/11

(iv) cosec C = 15/11

⇒ cosec C = (1/sin C)

⇒ cosec C = (Hypotenuse/Perpendicular)

⇒ cosec C = 15/11

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

Base = √(Hypotenuse)2 – (Perpendicular)2

⇒ Base = √(15)2 – (11)2

⇒ Base = √225 – 121

⇒ Base = √104

sin C = Perpendicular/Hypotenuse

⇒ sin C = 11/15

cos C = Base/Hypotenuse

⇒ cos C = √104/15

tan C = Perpendicular/Base

⇒ tan C = 11/√104

sec C = (1/cos C)

⇒ sec C = 15/√104

cot C = (1/tan C)

⇒ cot C = √104/11

(v) tan C = 5/12

tan C = Perpendicular/Base

tan C = 5/12

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ (Hypotenuse) = √(Perpendicular)2 + (Base)2

⇒ (Hypotenuse) = √(5)2 + (12)2

⇒ (Hypotenuse) = √25 + 144

⇒ (Hypotenuse) = √169

We get,

(Hypotenuse) = 13

cot C = (1/tan C)

⇒ cot C = 12/5

sin C = Perpendicular/Hypotenuse

⇒ sin C = 5/13

cos C = Base/Hypotenuse

⇒ cos C = 12/13

sec C = (1/cos C)

⇒ sec C = 13/12

cosec C = (1/sin C)

⇒ cosec C = 13/5

(vi) sin B = √3/2

sin B = Perpendicular/Hypotenuse

⇒ sin B = √3/2

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Base = √(Hypotenuse)2 – (Perpendicular)2

Base = √(2)2 – (√3)2

⇒ Base = √4 – 3

⇒ Base = √1

We get,

Base = 1

cos B = Base/Hypotenuse

⇒ cos B = 1/2

tan B = Perpendicular/Base

⇒ tan B = √3/1

⇒ tan B = √3

sec B = (1/Cos B)

⇒ sec B = 2

cot B = (1/tan B)

⇒ cot B = 1/√3

cosec B = 1/Sin A

⇒ cosec B = 2/√3

(vii) cos A = 7/25

cos A = Base/Hypotenuse

⇒ cos A = 7/25

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Perpendicular = √(Hypotenuse)2 – (Base)2

⇒ Perpendicular = √(25)2 – (7)2

⇒ Perpendicular = √625 – 49

⇒ Perpendicular = √576

We get,

Perpendicular = 24

sin A = Perpendicular/Hypotenuse

⇒ sin A = 24/25

tan A = Perpendicular/Base

sec A = (1/cos A)

⇒ sec A = 25/7

cot A = (1/tan A)

⇒ cot A = 7/24

cosec A = (1/sin A)

⇒ cosec A = 25/24

(viii) tan B = 8/15

tan B = Perpendicular/Base

tan B = 8/15

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ (Hypotenuse) = √(Perpendicular)2 + (Base)2

⇒ (Hypotenuse) = √(8)2 + (15)2

⇒ (Hypotenuse) = √64 + 225

⇒ (Hypotenuse) = √289

We get,

Hypotenuse = 17

⇒ cot B = 1/tan B

⇒ cot B = 15/8

sin B = Perpendicular/Hypotenuse

⇒ sin B =  8/17

cos B = Base/Hypotenuse

⇒ cos B = 15/17

sec B = (1/cos B)

⇒ sec B = 17/15

cosec B = (1/sin B)

⇒ cosec B = 17/8

(ix) sec B = 15/12

sec B = (1/cos B)

⇒ sec B = Hypotenuse/Base

⇒ sec B = 15/12

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Perpendicular = √(Hypotenuse)2 – (Base)2

⇒ Perpendicular = √(15)2 – (12)2

⇒ Perpendicular = √225 – 144

⇒ Perpendicular = √81

We get,

Perpendicular = 9

sin B = Perpendicular/Hypotenuse

⇒ sin B = 9/15

tan B = Perpendicular/Base

⇒ tan B = 9/12

cot B = (1/tan B)

⇒ cot B = 12/9

cosec B = (1/sin B)

⇒ cosec B = 15/9

cos B = Base/Hypotenuse

⇒ cos B = 12/15

(x) cosec C = √10

⇒ cosec C = (1/sin C)

cosec C = Hypotenuse/Perpendicular

⇒ cosec C = √10/1

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

Base = (Hypotenuse)2 – (Perpendicular)2

⇒ Base = (10)2 – (1)2

⇒ Base = 10 – 1

⇒ Base = 9

We get,

Base = 3

sin C = Perpendicular/Hypotenuse

⇒ sin C = 1/10

cos C = Base/Hypotenuse

⇒ cos C = 3/10

tan C = Perpendicular/Base

⇒ tan C = 1/3

sec C = (1/cos C)

⇒ sec C = 10/3

cot C = (1/tan C)

⇒ cot C = 3

2. In ABC, A = 90°. If AB = 5 units and AC = 12 units, find:

(i) sin B

(ii) cos C

(iii) tan B

In △ABC,

BC2 = AB2 + AC2

⇒ BC = √AB2 + AC2

⇒ BC = √52 + 122

⇒ BC = 25 + 144

⇒ BC = 169

We get,

BC = 13

AC = 12 units

BC = 13 units

AB = 5 units

(i) sin B = Perpendicular/Hypotenuse

sin B = AC/BC

⇒ sin B = 12/13

(ii) cos C = Base/Hypotenuse

⇒ cos C = AC/BC

⇒ cos C = 12/13

(iii) tan B = Perpendicular/Base

⇒ tan B = AC/AB

⇒ tan B = 12/5

3. In ABC, B = 90°. If AB = 12 units and BC = 5 units, find:

(i) sin A

(ii) tan A

(iii) cos C

(iv) cot C

In △ABC,

AC2 = AB2 + BC2

⇒ AC = 122 + 52

⇒ AC = 144 + 25

⇒ AC = 169

We get,

AC = 13

AB = 12 units

BC = 5 units

AC = 13 units

(i) sin A = Perpendicular/Hypotenuse

⇒ sin A = BC/AC

⇒ sin A = 5/13

(ii) tan A = Perpendicular/Base

⇒ tan A = BC/AB

⇒ tan A = 5/12

(iii) cos C = Base/Hypotenuse

⇒ cos C = BC/AC

⇒ cos C = 5/13

(iv) cot C = Base/Perpendicular

⇒ cot C = BC/AB

⇒ cot C = 5/12

4. If sin A = 3/5, find cos A and tan A.

Given

sin A = 3/5

sin A = Perpendicular/Hypotenuse

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ (Base)2 = (Hypotenuse)2 – (Perpendicular)2

⇒ Base = √(Hypotenuse)2 – (Perpendicular)2

⇒ Base = √52 – 32

⇒ Base = √25 – 9

⇒ Base = √16

We get,

Base = 4

cos A = Base/Hypotenuse

⇒ cos A = 4/5

tan A = Perpendicular/Base

⇒ tan A = 3/4

5. If sin Î¸ = 8/17, find the other five trigonometric ratios.

Given

sin Î¸ = 8/17

⇒ sin Î¸ = Perpendicular/Hypotenuse

Base = √(Hypotenuse)2 – (Perpendicular)2

⇒ Base = √172 – 82

⇒ Base = √289 – 64

⇒ Base = √225

We get,

Base = 15

cos Î¸ = Base/Hypotenuse

⇒ cos Î¸ = 15/17

tan Î¸ = Perpendicular/Base

⇒ tan Î¸ = 8/15

cosec Î¸ = 1/sin Î¸

⇒ cosec Î¸ = 17/8

sec Î¸ = 1/cos Î¸

⇒ sec Î¸ = 17/15

cot Î¸ = 1/tan Î¸

⇒ cot Î¸ = 15/8

6. If tan A = 0.75, find the other trigonometric ratios for A.

Given,

tan A = 0.75

⇒ tan A = 75/100

We get,

tan A = 3/4

tan A = Perpendicular/Base

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Hypotenuse = √(Perpendicular)2 + (Base)2

⇒ Hypotenuse = √32 + 42

⇒ Hypotenuse = √9 + 16

⇒ Hypotenuse = √25

We get,

Hypotenuse = 5

sin A = Perpendicular/Hypotenuse

⇒ sin A = 3/5

⇒ sin A = 0.6

cos A = Base/Hypotenuse

⇒ cos A = 4/5

⇒ cos A = 0.8

cosec A = 1/sin A

⇒ cosec A = 5/3

⇒ cosec A = 1.66

sec A = 1/cos A

⇒ sec A = 5/4

⇒ sec A = 1.25

cot A = 1/tan A

⇒ cot A = 4/3

⇒ cot A = 1.33

7. If sin A = 0.8, find the other trigonometric ratios for A.

Given

sin A = 0.8

⇒ sin A = 8/10

⇒ sin A = 4/5

sin A = Perpendicular/Hypotenuse

By Pythagoras theorem,

We have

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Base =√(Hypotenuse)2 – (Perpendicular)2

⇒ Base = √52 – 42

⇒ Base = √25 – 16

⇒ Base = √9

We get,

Base = 3

cos A = Base/Hypotenuse

⇒ cos A = 3/5

⇒ cos A = 0.6

tan A = Perpendicular/Base

⇒ tan A = 4/3

⇒ tan A = 1.33

cosec A = 1/sin A

⇒ cosec A = 5/4

⇒ cosec A = 1.25

sec A = 1/cos A

⇒ sec A = 5/3

⇒ sec A = 1.66

cot A = 1/tan A

⇒ cot A = 3/4

⇒ cot A = 0.75

8. If 8 tan Î¸ = 15, find

(i) sin Î¸

(ii) cot Î¸

(iii) sin2 Î¸ – cot2 Î¸

Given

8 tanÎ¸ = 15

⇒ tan Î¸ = 15/8

⇒ tan Î¸ = Perpendicular/Base

By Pythagoras theorem,

We have,

(Hypotenuse)2 = (Perpendicular)2 + (Base)2

⇒ Hypotenuse = √(Perpendicular)2 + (Base)2

⇒ Hypotenuse = √152 + 82

⇒ Hypotenuse = √225 + 64

⇒ Hypotenuse = √289

We get,

Hypotenuse = 17

(i) sin Î¸ = Perpendicular/Hypotenuse

⇒ sin Î¸ = 15/17

(ii) cot Î¸ = 1/tan Î¸

⇒ cot Î¸ = 8/15

(iii) sin2 Î¸ – cot2 Î¸ = (sin Î¸ + cot Î¸) (sin Î¸ – cot Î¸)

⇒ sin2 Î¸ – cot2 Î¸ = {(15/17) + (8/15)} {(15/17) – (8/15)}

⇒ sin2 Î¸ – cot2 Î¸ = {(225 + 136)/255}{(225 – 136)/255}

⇒ sin2 Î¸ – cot2 Î¸ = (361/255) (89/255)

On calculation, we get,

sin2 Î¸ – cot2 Î¸ = 32129/65025

9. In an isosceles triangle ABC, AB = BC = 6 cm and B = 90°. Find the values of

(a) cos C

(b) cosec C

(c) cos2 C + cosec2 C

△ABC is an isosceles right-angled triangle

Therefore,

AC2 = AB2 + BC2

⇒ AC2 = 62 + 62

⇒ AC2 = 36 + 36

⇒ AC2 = 72

We get,

⇒ AC = 6√2 cm

(a) cos C = BC/AC

⇒ cos C = 6/6√2

⇒ cos C = 1/√2

(b) cosec C = AC/AB

⇒ cosec C = 6√2/6

⇒ cosec C = √2

(c) cos2 C + cosec2 C = (1/√2)2 + (√2)2

⇒ cos2 C + cosec2 C = (1/2) + 2

On further calculation, we get,

cos2 C + cosec2 C = 5/2

10. In the given figure, AD is the median on BC from A. If AD = 8 cm and BC = 12 cm, find the value of

(a) sin x

(b) cos y

(c) tan x. cot y

(d) (1/sin2 x) – (1/ tan2 x)

Since, AD is the median on BC,

We have,

BD = DC = (1/2)× BC

= (1/2) ×12

= 6 cm

△ADB is a right angled triangle

Therefore,

⇒ AB2 = 82 + 62

⇒ AB2 = 64 + 36

⇒ AB2 = 100

We get,

AB = 10 cm

△ADC is a right angled triangle

Therefore,

⇒ AC2 = 82 + 62

⇒ AC2 = 64 + 36

⇒ AC2 = 100

We get,

AC = 10 cm

⇒ sin x = 8/10

⇒ sin x = 4/5

⇒ cos y = 8/10

⇒ cos y = 4/5

(c) cos x = BD/AB

⇒ cos x = 6/10

⇒ cos x = 3/5

And,

sin y = DC/AC

⇒ sin y = 6/10

⇒ sin y = 3/5

Hence,

tan x = sin x/cos x

⇒ tan x = (4/5)/(3/5)

We get,

tan x = 4/3

⇒ cot y = cos y/sin y

⇒ cot y = (4/5)/(3/5)

We get,

cot y = 4/3

Therefore,

tan x. cot y = (4/3) ×(4/3)

⇒ tan x . cot y = 16/9

(d) (1/sin2 x) – (1/tan2 x) = 1/(4/5)2 – 1/(4/3)2

On calculating further, we get,

(1/sin2 x ) – (1/tan2 x) = (25/16) – (9/16)

⇒ (1/sin2 x) – (1/ tan2 x) = (16/16)

⇒ (1/sin2 x) – (1/tan2 x) = 1

11In a right- angled triangle PQR, PQR = 90°, QS PR and tan R = (5/12), find the value of

(a) sin PQS

(b) tan SQR

tan R = 5/12

PQ/QR = 5/12

Hence,

PQ = 5 and QR = 12

In right-angled △PQR,

PR2 = PQ2 + QR2

⇒ PR2 = 52 + 122

⇒ PR2 = 25 + 144

⇒ PR2 = 169

⇒ PR = √169

We get,

PR = 13

(a) ∠PQS + ∠P = 90° and ∠P + ∠R = 90°

Hence,

∠PQS + ∠P = ∠P + ∠R

⇒ ∠PQS = ∠R

Therefore,

sin ∠PQS = sin R = PQ/PR = 5/13

(b) ∠SQR + ∠R = 90° and ∠R + ∠P = 90°

Hence,

∠SQR + ∠R = ∠R + ∠P

⇒ ∠SQR = ∠P

Therefore,

tan ∠SQR = tan P = QR/PQ = 12/5

12. In the given figure, ABC is right angled at B. AD divides BC in the ratio 1: 2.

Find

Given

△ABC is right angled at B

BD: DC = 1: 2 as AD divides BC in the ratio 1: 2

i.e BD = x, DC = 2x

Hence,

BC = 3x

(i) tan ∠BAC/tan ∠BAD = (BC/AB)/(BD/AB)

⇒ tan ∠BAC/tan ∠BAD = BC/BD

⇒ tan ∠BAC/tan ∠BAD = 3x/x

⇒ tan ∠BAC/tan ∠BAD = 3

(ii) cot ∠BAC/cot ∠BAD = (AB/BC)/(AB/BD)

⇒ cot ∠BAC/cot ∠BAD = BD/BC

⇒ cot ∠BAC/cot ∠BAD = x/3x

⇒ cot ∠BAC/cot ∠BAD = 1/3

13. If sin A = 7/25, find the value of:

(a) (2 tan A)/(cot A – sin A)

(b) cos A + (1/cot A)

(c) cot2 A – cosec2 A

Consider △ABC, where ∠B = 90°

sin A = Perpendicular/Hypotenuse

⇒ sin A = BC/AC

⇒ sin A = 7/25

cosec A = 1/sin A

⇒ cosec A = 25 7

By Pythagoras theorem,

We have,

AC2 = AB2 + BC2

⇒ AB2 = AC2 – BC2

⇒ AB2 = 252 – 72

⇒ AB2 = 625 – 49

⇒ AB2 = 576

⇒ AB = √576

We get,

AB = 24

Now,

cos A = Base/Hypotenuse

⇒ cos A = AB/AC

⇒ cos A = 24/25

tan A = Perpendicular/Base

⇒ tan A = BC/AB

⇒ tan A = 7/24

cot A = (1/tan A)

(a) 2 tan A/cot A – sin A = {2×(7/24)}/(24/7 – 7/25)

On further calculation, we get,

= (7/12)/(551/175)

= (7/12)×(175/551)

We get,

= 1225/6612

(b) cos A + 1/cot A = cos A + tan A

⇒ cos A + 1/cot A = (24/25) + (7/24)

On calculating further, we get,

cos A + 1/cot A = (576 + 175)/600

⇒ cos A + 1/cot A = 751/600

(c) cot2 A – cosec2 A = (24/7)2 – (25/7)2

⇒ cot2 A – cosec2 A = (576/49) – (625/49)

⇒ cot2 A – cosec2 A = (576 – 625)/49

We get,

cot2 A – cosec2 A = – 49/49

⇒ cot2 A – cosec2 A = – 1

14. If cosec Î¸ = 29/20, find the value of:

(a) cosec Î¸ – (1/cot Î¸)

(b) sec Î¸/(tan Î¸ – cosec Î¸)

Consider △ABC, where ∠A = 90°

cosec Î¸ = Hypotenuse/Perpendicular

⇒ cosec Î¸ = BC/AB

⇒ cosec Î¸ = 29/20

By Pythagoras theorem,

We have,

BC2 = AB2 + AC2

⇒ AC2 = BC2 – AB2

⇒ AC2 = 292 – 202

⇒ AC2 = 841 – 400

⇒ AC2 = 441

⇒ AC = √441

We get,

AC = 21

Now,

sec Î¸ = Hypotenuse/Base

⇒ sec Î¸ = BC/AC

⇒ sec Î¸ = 29/21

tan Î¸ = Perpendicular/Base

⇒ tan Î¸ = AB/AC

⇒ tan Î¸ = 20/21

cot Î¸ = 1/tan Î¸

⇒ cot Î¸ = 21/20

(a) cosec Î¸ – (1/cot Î¸) = (29/20) – {1/(21/20)}

⇒ cosec Î¸ – (1/cot Î¸) = (29/20) – (20/21)

⇒ cosec Î¸ – (1/cot Î¸) = (609 – 400)/420

We get,

cosec Î¸ – (1/cot Î¸) = 209/420

(b) sec Î¸/(tan Î¸ – cosec Î¸) = (29/21)/(20/21 – 29/20)

⇒ sec Î¸/(tan Î¸ – cosec Î¸) = (29/21)/{(400 – 609)/420}

⇒ sec Î¸/(tan Î¸ – cosec Î¸) = (29/21)/{(-209/420)}

⇒ sec Î¸/(tan Î¸ – cosec Î¸) = (29/21)×(-420/209)

⇒ sec Î¸/(tan Î¸ – cosec Î¸) = –580/209

15. In the given figure, AC = 13 cm, BC = 12 cm and B = 90°. Without using tables, find the values of:

(a) sin A cos A

(b) (cos A – sin A) / (cos A + sin A)

△ABC is a right- angled triangle.

By Pythagoras theorem,

We have,

AC2 = AB2 + BC2

⇒ AB2 = AC2 – BC2

⇒ AB2 = 132 – 122

⇒ AB2 = 169 – 144

⇒ AB2 = 25

⇒ AB = √25

We get,

AB = 5 cm

sin A = BC/AC

⇒ sin A = 12/13

cos A = AB/AC

⇒ cos A = 5/13

(a) sin A cos A = (12/13)× (5/13)

⇒ sin A cos A = 60/169

(b) (cos A – sin A)/(cos A + sin A) = {(5/13) – (12/13)}/{(5/13) + (12/13)}

⇒ (cos A – sin A)/(cos A + sin A) = (-7/13)/(17/13)

⇒ (cos A – sin A)/(cos A + sin A) = (-7/13) x (13/17)

⇒ (cos A – sin A)/(cos A + sin A) = -7/17

16. In the given figure, PQR is a triangle, in which QS PR, Qs = 3 cm, PS = 4 cm and QR = 12 cm, find the value of:

(a) sin P

(b) cot2 P - cosec2 P

(c) 4sin2 R – 1/tan2 P

(a)

(b)
(c)

17. In tan Î¸ = 1, find the value of 5 cot2 Î¸ + sin2 Î¸ - 1.

18. In the given figure, Q = 90°, PS is a median on QR from P, and RT divides PQ in the ratio 1 : 2. Find :

(i) (tan PSQ)/(tan PRQ)

(ii) (tan TSQ)/(tan PRQ)

(i)

(ii)

19. In the given figure, AD is perpendicular to BC. Find:

(a) 5 cos x

(b) 15 tan y

(c) 5 cos x – 12 sin y + tan x

(d) 3/sinx + 4/cos y – tan y

(a)

(b)
(c)
(d)

20. In a right-angled triangle ABC, B = 90°, BD = 3, DC = 4, and AC = 13. A point D is inside the triangle such as BDC = 90°.

Find the values of

(a) 2 tan BAC – sin BCD

(b) 3 – 2 cos BAC + 3 cot BCD

(a)

(b)

21. If cos 24 Î¸ – 7 sin Î¸, find sin Î¸ + cos Î¸.

22. If 4 sin Î¸ = 3 cos Î¸, find

(a) tan2 Î¸ + cot2 Î¸

(b) (6 sin Î¸ - 2 cos Î¸)/(6 sin Î¸ + 2 cos Î¸)

(a)

(b)

23. If 8 tan A = 15, find sin A – cos A.

24. If 3 cos Î¸ - 4 sin Î¸ = 2 cos Î¸ + sin Î¸, find tan Î¸.

25. If cos Î¸ = 3, find the value of (4 cos Î¸ – sin Î¸)/(2 cos Î¸ + sin Î¸).

26. If sin 4 Î¸ = √13,find the value of

(a) (4 sin Î¸ – 3 cos Î¸)/(2 sin Î¸ + 6 cos Î¸)

(b) (4 sin3 Î¸ – 3 sin Î¸)

(a)

(b)

27. If 5 tan Î¸ = 12, find the value of (2 sin Î¸ - 3 cos Î¸)/(4 sin Î¸ – 9 cos Î¸).

28. If cot Î¸ = 1/√3, show that (1 – cos2 Î¸)/(2 – sin2 Î¸) = 3/5

29. If cot Î¸ = 1/√3, show that (1 – cos2 Î¸)/(2 – sin2 Î¸) = 3/5

30. If cosec Î¸ = 1.(9/20), show that (1 – sin Î¸ + cos Î¸)/(1 + sin Î¸ + cos Î¸) = 3/7

31. If b tan Î¸ = a, find the values of (cos Î¸ + sin Î¸)/(cos Î¸ – sin Î¸).

32. If a cot Î¸ = b, prove that (a sin Î¸ – b cos Î¸)/(a sin Î¸ + b cos Î¸) = (a2 – b2)/(a2 + b2)

33. If cot Î¸ = b, prove that (a sin Î¸ – b cos Î¸)/(a sin Î¸ + b cos Î¸) = (a2 – b2)/(a2 + b2)

34. If cot Î¸ = √7, show that (cosec2 Î¸ – sec2 Î¸)/(cosec2 Î¸ + sec2 Î¸) = 3/4

35. If 12 cosec Î¸ = 13, find the value of (sin2 Î¸ – cos2 Î¸)/(2 sin Î¸ cos Î¸) × 1/tan2 Î¸.

36. If 12 cos Î¸ = 13, find the value of (2 sin Î¸ cos Î¸)/(cos2 Î¸ – sin2 Î¸).

37. If sec A = 5/4, verify that (3 sin A – 4 sin3 A)/(4 cos3 A – 3 cos A) = (3 tan A – tan3 A)/(1 – 3 tan2 A).

38. If sin Î¸ = 3/4, prove that √[(cosec2Î¸ – cot2Î¸)/(sec2Î¸-1) = √7/3

39. If sec A = 17/8, verify that (3 – 4 sin2 A)/(4 cos2 A – 3) = (3 – tan2 A)/ |(1 – 3 tan2 A).