Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios

Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios provide solutions in a stepwise manner as per students’ intelligence levels. The main aim of developing these solutions is to make the learning process smooth and hassle-free for students. Those who aspire to excel in solving difficult problems easily can practise Frank Solutions regularly.

Chapter 26, Trigonometrical Ratios, contains problems pertaining to the measurement of the sides and angles of a triangle. The branch of Mathematics which deals with the study of the sides and angles of a right-angled triangle is known as Trigonometry. It is essential to understand this chapter in-depth, as it plays a vital role in higher classes also. Students can download the Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios PDF from the links provided below.

Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios Download PDF

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Access Frank Solutions for Class 9 Maths Chapter 26 Trigonometrical Ratios

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

Solution:

(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 = 900. If AB = 5 units and AC = 12 units, find:

(i) sin B

(ii) cos C

(iii) tan B

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 1

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 = 900. If AB = 12 units and BC = 5 units, find:

(i) sin A

(ii) tan A

(iii) cos C

(iv) cot C

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 2

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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

Solution:

Given

8 tan θ = 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 = 900. Find the values of

(a) cos C

(b) cosec C

(c) cos2 C + cosec2 C

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 3

△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)

FRANK Solutions Class 9 Maths Chapter 26 - 4

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 5

Since AD is the median on BC,

We have,

BD = DC = (1/2) x BC

= (1/2) x 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) x (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 = 900, QS ⊥PR and tan R = (5/ 12), find the value of

(a) sin ∠PQS

(b) tan ∠SQR

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 6

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 = 900 and ∠P + ∠R = 900

Hence,

∠PQS + ∠P = ∠P + ∠R

∠PQS = ∠R

Therefore,

sin ∠PQS = sin R = PQ / PR = 5 / 13

(b) ∠SQR + ∠R = 900 and ∠R + ∠P = 900

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

FRANK Solutions Class 9 Maths Chapter 26 - 7

Solution:

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

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 8

Consider △ABC, where ∠B = 900

sin A = Perpendicular / Hypotenuse

sin A = BC / AC

sin A = 7/25

cosec A = 1/sin A

cosec A = 25 / 7

By the 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 x (7/24)} / {(24/7) – (7/25)}

On further calculation, we get,

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

= (7/12) x (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)

Solution:

FRANK Solutions Class 9 Maths Chapter 26 - 9

Consider △ABC, where ∠A = 900

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) x (-420 / 209)

sec θ / (tan θ – cosec θ) = – 580 / 209

15. In the given figure, AC = 13 cm, BC = 12 cm and ∠B = 900. Without using tables, find the values of:

(a) sin A cos A

(b) (cos A – sin A) / (cos A + sin A)

FRANK Solutions Class 9 Maths Chapter 26 - 10

Solution:

△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) x (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

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