Frank Chapter 26 Trigonometric Ratios ICSE Solutions Class 9 Math

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

Answer

(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

⇒ tan A = 24/7

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

Answer

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

Answer

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.

Answer

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 θ = 8/17, find the other five trigonometric ratios.

Answer

Given

sin θ = 8/17

⇒ sin θ = Perpendicular/Hypotenuse

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

⇒ Base = √17² – 8²

⇒ Base = √289 – 64

⇒ Base = √225

We get,

Base = 15

cos θ = Base/Hypotenuse

⇒ cos θ = 15/17

tan θ = Perpendicular/Base

⇒ tan θ = 8/15

cosec θ = 1/sin θ

⇒ cosec θ = 17/8

sec θ = 1/cos θ

⇒ sec θ = 17/15

cot θ = 1/tan θ

⇒ cot θ = 15/8

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

Answer

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.

Answer

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 θ = 15, find

(i) sin θ

(ii) cot θ

(iii) sin² θ – cot² θ

Answer

Given

8 tanθ = 15

⇒ tan θ = 15/8

⇒ tan θ = 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 θ = Perpendicular/Hypotenuse

⇒ sin θ = 15/17

(ii) cot θ = 1/tan θ

⇒ cot θ = 8/15

(iii) sin² θ – cot² θ = (sin θ + cot θ) (sin θ – cot θ)

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

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

⇒ sin² θ – cot² θ = (361/255) (89/255)

On calculation, we get,

sin² θ – cot² θ = 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

Answer

△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)

Answer

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 = AD² + BD²

⇒ AB² = 8² + 6²

⇒ AB² = 64 + 36

⇒ AB² = 100

We get,

AB = 10 cm

△ADC is a right angled triangle

Therefore,

AC² = AD² + DC²

⇒ AC² = 8² + 6²

⇒ AC² = 64 + 36

⇒ AC² = 100

We get,

AC = 10 cm

(a) sin x = AD/AB

⇒ sin x = 8/10

⇒ sin x = 4/5

(b) cos y = AD/AC

⇒ 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

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

(a) sin ∠PQS

(b) tan ∠SQR

Answer

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

(i) tan ∠BAC/tan ∠BAD

(ii) cot ∠BAC/cot ∠BAD

Answer

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

Answer

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)

⇒ cot A = 24/7

(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 θ = 29/20, find the value of:

(a) cosec θ – (1/cot θ)

(b) sec θ/(tan θ – cosec θ)

Answer

Consider △ABC, where ∠A = 90°

cosec θ = Hypotenuse/Perpendicular

⇒ cosec θ = BC/AB

⇒ cosec θ = 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 θ = Hypotenuse/Base

⇒ sec θ = BC/AC

⇒ sec θ = 29/21

tan θ = Perpendicular/Base

⇒ tan θ = AB/AC

⇒ tan θ = 20/21

cot θ = 1/tan θ

⇒ cot θ = 21/20

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

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

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

We get,

cosec θ – (1/cot θ) = 209/420

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

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

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

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

⇒ sec θ/(tan θ – cosec θ) = –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)

Answer

△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

Answer

(a)

(b)

(c)

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

Answer

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)

Answer

(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

Answer

(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

Answer

(a)

(b)

21. If cos 24 θ – 7 sin θ, find sin θ + cos θ.

Answer

22. If 4 sin θ = 3 cos θ, find

(a) tan² θ + cot² θ

(b) (6 sin θ - 2 cos θ)/(6 sin θ + 2 cos θ)

Answer

(a)

(b)

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

Answer

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

Answer

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

Answer

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

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

(b) (4 sin³ θ – 3 sin θ)

Answer

(a)

(b)

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Answer

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).

Answer

38. If sin θ = 3/4, prove that √[(cosec²θ – cot²θ)/(sec²θ-1) = √7/3

Answer

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

Answer

40. If tan θ = 4, prove that √[(se²θ – cosecθ)/(secθ + cosecθ) = 1/√7

Answer

41. If tan θ = m/n, show that (m sin θ – n cos θ)/(m sin θ + n cos θ) = (m² – n²)/(m² + n²).

Answer