# 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 studentsâ€™ 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.

## 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 = âˆš3 / 2

(vii) cos A = 7 / 25

(viii) tan B = 8 / 15

(ix) sec B = 15 / 12

(x) cosec C = âˆš10

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 = âˆš(Hypotenuse)2 â€“ (Perpendicular)2

Base = âˆš(13)2 â€“ (12)2

Base = âˆš169 â€“ 144

Base = âˆš25

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 = âˆš(Hypotenuse)2 â€“ (Base)2

Perpendicular = âˆš(5)2 â€“ (4)2

Perpendicular = âˆš25 â€“ 16

Perpendicular = âˆš9

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) = âˆš(Perpendicular)2 + (Base)2

(Hypotenuse) = âˆš(11)2 + (1)2

(Hypotenuse) = âˆš(121 + 1)

(Hypotenuse) = âˆš122

Cos A = Base / Hypotenuse

Cos A = 1 / âˆš122

Tan A = Perpendicular / Base

Tan A = 11

Sec A = (1 / cos A)

Sec A = âˆš122

Sin A = Perpendicular / Hypotenuse

Sin A = 11 / âˆš122

Cosec A = (1 / sin A)

Cosec A = âˆš122 / 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 = âˆš(Hypotenuse)2 â€“ (Perpendicular)2

Base = âˆš(15)2 â€“ (11)2

Base = âˆš225 â€“ 121

Base = âˆš104

Sin C = Perpendicular / Hypotenuse

Sin C = 11 / 15

Cos C = Base / Hypotenuse

Cos C = âˆš104 / 15

Tan C = Perpendicular / Base

Tan C = 11 / âˆš104

Sec C = (1 / Cos C)

Sec C = 15 / âˆš104

Cot C = (1 / tan C)

Cot C = âˆš104 / 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) = âˆš(Perpendicular)2 + (Base)2

(Hypotenuse) = âˆš(5)2 + (12)2

(Hypotenuse) = âˆš25 + 144

(Hypotenuse) = âˆš169

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 = âˆš3 / 2

Sin B = Perpendicular / Hypotenuse

Sin B = âˆš3 / 2

By Pythagoras theorem,

We have,

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

Base = âˆš(Hypotenuse)2 â€“ (Perpendicular)2

Base = âˆš(2)2 â€“ (âˆš3)2

Base = âˆš4 â€“ 3

Base = âˆš1

We get,

Base = 1

Cos B = Base / Hypotenuse

Cos B = 1 / 2

Tan B = Perpendicular / Base

Tan B = âˆš3 / 1

Tan B = âˆš3

Sec B = (1 / Cos B)

Sec B = 2

Cot B = (1 / tan B)

Cot B = 1 / âˆš3

Cosec B = 1 / Sin A

Cosec B = 2 / âˆš3

(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 = âˆš(Hypotenuse)2 â€“ (Base)2

Perpendicular = âˆš(25)2 â€“ (7)2

Perpendicular = âˆš625 â€“ 49

Perpendicular = âˆš576

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) = âˆš(Perpendicular)2 + (Base)2

(Hypotenuse) = âˆš(8)2 + (15)2

(Hypotenuse) = âˆš64 + 225

(Hypotenuse) = âˆš289

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 = âˆš(Hypotenuse)2 â€“ (Base)2

Perpendicular = âˆš(15)2 â€“ (12)2

Perpendicular = âˆš225 â€“ 144

Perpendicular = âˆš81

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 = âˆš10

Cosec C = (1 / sin C)

Cosec C = Hypotenuse / Perpendicular

Cosec C = âˆš10 / 1

By Pythagoras theorem,

We have,

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

Base = âˆš(Hypotenuse)2 â€“ (Perpendicular)2

Base = âˆš(âˆš10)2 â€“ (1)2

Base = âˆš10 â€“ 1

Base = âˆš9

We get,

Base = 3

Sin C = Perpendicular / Hypotenuse

Sin C = 1 / âˆš10

Cos C = Base / Hypotenuse

Cos C = 3 / âˆš10

Tan C = Perpendicular / Base

Tan C = 1 / 3

Sec C = (1 / cos C)

Sec C = âˆš10 / 3

Cot C = (1 / tan C)

Cot C = 3

2. In â–³ABC, âˆ A = 900. If AB = 5 units and AC = 12 units, find:

(i) sin B

(ii) cos C

(iii) tan B

Solution:

In â–³ABC,

BC2 = AB2 + AC2

BC = âˆšAB2 + AC2

BC = âˆš52 + 122

BC = âˆš25 + 144

BC = âˆš169

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 â–³ABC, âˆ B = 900. If AB = 12 units and BC = 5 units, find:

(i) sin A

(ii) tan A

(iii) cos C

(iv) cot C

Solution:

In â–³ABC,

AC2 = AB2 + BC2

AC = âˆš122 + 52

AC = âˆš144 + 25

AC = âˆš169

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 â€“ (Perpendicular)2

Base = âˆš(Hypotenuse)2 â€“ (Perpendicular)2

Base = âˆš52 – 32

Base = âˆš25 â€“ 9

Base = âˆš16

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.

Solution:

Given

sin Î¸ = 8 / 17

sin Î¸ = Perpendicular / Hypotenuse

Base = âˆš(Hypotenuse)2 â€“ (Perpendicular)2

Base = âˆš172 – 82

Base = âˆš289 â€“ 64

Base = âˆš225

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.

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 = âˆš(Perpendicular)2 + (Base)2

Hypotenuse = âˆš32 + 42

Hypotenuse = âˆš9 + 16

Hypotenuse = âˆš25

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 = âˆš(Hypotenuse)2 â€“ (Perpendicular)2

Base = âˆš52 – 42

Base = âˆš25 â€“ 16

Base = âˆš9

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¸ â€“ cot2 Î¸

Solution:

Given

8 tan Î¸ = 15

tan Î¸ = 15 / 8

tan Î¸ = Perpendicular / Base

By Pythagoras theorem,

We have,

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

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

Hypotenuse = âˆš152 + 82

Hypotenuse = âˆš225 + 64

Hypotenuse = âˆš289

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¸ â€“ cot2 Î¸ = (sin Î¸ + cot Î¸) (sin Î¸ â€“ cot Î¸)

sin2 Î¸ â€“ cot2 Î¸ = {(15 / 17) + (8 / 15)} {(15 / 17) â€“ (8 / 15)}

sin2 Î¸ â€“ cot2 Î¸ = {(225 + 136) / 255}{(225 â€“ 136) / 255}

sin2 Î¸ â€“ cot2 Î¸ = (361 / 255) (89 / 255)

On calculation, we get,

sin2 Î¸ â€“ cot2 Î¸ = 32129 / 65025

9. In an isosceles triangle ABC, AB = BC = 6 cm and âˆ B = 900. Find the values of

(a) cos C

(b) cosec C

(c) cos2 C + cosec2 C

Solution:

â–³ABC is an isosceles right-angled triangle

Therefore,

AC2 = AB2 + BC2

AC2 = 62 + 62

AC2 = 36 + 36

AC2 = 72

We get,

AC = 6âˆš2 cm

(a) cos C = BC / AC

cos C = (6 / 6âˆš2)

cos C = (1 / âˆš2)

(b) cosec C = AC / AB

cosec C = (6âˆš2 / 6)

cosec C = âˆš2

(c) cos2 C + cosec2 C = (1/âˆš2)2 + (âˆš2)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) â€“ (1/ tan2 x)

Solution:

Since AD is the median on BC,

We have,

BD = DC = (1/2) x BC

= (1/2) x 12

= 6 cm

Therefore,

AB2 = 82 + 62

AB2 = 64 + 36

AB2 = 100

We get,

AB = 10 cm

Therefore,

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) â€“ (1/ tan2 x) = {1/ (4/5)2} â€“ {1/ (4/3)2}

On calculating further, we get,

(1/ sin2 x ) â€“ (1/ tan2 x) = (25 / 16) â€“ (9 / 16)

(1/ sin2 x) â€“ (1/ tan2 x) = (16 / 16)

(1/ sin2 x) â€“ (1/ tan2 x) = 1

11. In a right-angled triangle PQR, âˆ PQR = 900, QS âŠ¥PR and tan R = (5/ 12), find the value of

(a) sin âˆ PQS

(b) tan âˆ SQR

Solution:

tan R = 5 / 12

PQ / QR = 5 / 12

Hence,

PQ = 5 and QR = 12

In right-angled â–³PQR,

PR2 = PQ2 + QR2

PR2 = 52 + 122

PR2 = 25 + 144

PR2 = 169

PR = âˆš169

We get,

PR = 13

(a) âˆ PQS + âˆ P = 900 and âˆ P + âˆ R = 900

Hence,

âˆ PQS + âˆ P = âˆ P + âˆ R

âˆ PQS = âˆ R

Therefore,

sin âˆ PQS = sin R = PQ / PR = 5 / 13

(b) âˆ SQR + âˆ R = 900 and âˆ R + âˆ P = 900

Hence,

âˆ SQR + âˆ R = âˆ R + âˆ P

âˆ SQR = âˆ P

Therefore,

tan âˆ SQR = tan P = QR / PQ = 12 / 5

12. In the given figure, â–³ABC is right-angled at B. AD divides BC in the ratio 1: 2.

Find

(i) tan âˆ BAC / tan âˆ BAD

(ii) cot âˆ BAC / cot âˆ BAD

Solution:

Given

â–³ABC 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 âˆ BAC / tan âˆ BAD = (BC / AB) / (BD / AB)

tan âˆ BAC / tan âˆ BAD = BC / BD

tan âˆ BAC / tan âˆ BAD = 3x / x

tan âˆ BAC / tan âˆ BAD = 3

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

cot âˆ BAC / cot âˆ BAD = BD / BC

cot âˆ BAC / cot âˆ BAD = x / 3x

cot âˆ BAC / cot âˆ BAD = 1 / 3

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

(a) (2 tan A) / (cot A â€“ sin A)

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

(c) cot2 A â€“ cosec2 A

Solution:

Consider â–³ABC, where âˆ 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 â€“ BC2

AB2 = 252 – 72

AB2 = 625 â€“ 49

AB2 = 576

AB = âˆš576

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 â€“ sin A = {2 x (7/24)} / {(24/7) â€“ (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 â€“ cosec2 A = (24 / 7)2 â€“ (25 / 7)2

cot2 A â€“ cosec2 A = (576 / 49) â€“ (625 / 49)

cot2 A â€“ cosec2 A = (576 â€“ 625) / 49

We get,

cot2 A â€“ cosec2 A = – 49 / 49

cot2 A â€“ cosec2 A = – 1

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

(a) cosec Î¸ â€“ (1/ cot Î¸)

(b) sec Î¸ / (tan Î¸ â€“ cosec Î¸)

Solution:

Consider â–³ABC, where âˆ A = 900

Cosec Î¸ = Hypotenuse / Perpendicular

Cosec Î¸ = BC / AB

Cosec Î¸ = 29 / 20

By Pythagoras theorem,

We have,

BC2 = AB2 + AC2

AC2 = BC2 â€“ AB2

AC2 = 292 â€“ 202

AC2 = 841 â€“ 400

AC2 = 441

AC = âˆš441

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¸ â€“ (1/ cot Î¸) = (29 / 20) â€“ {1 / (21/20)}

cosec Î¸ â€“ (1/ cot Î¸) = (29 / 20) â€“ (20 / 21)

cosec Î¸ â€“ (1/ cot Î¸) = (609 â€“ 400) / 420

We get,

cosec Î¸ â€“ (1/ cot Î¸) = 209 / 420

(b) sec Î¸ / (tan Î¸ â€“ cosec Î¸) = (29 / 21) / {(20 / 21) â€“ (29 / 20)}

sec Î¸ / (tan Î¸ â€“ cosec Î¸) = (29 / 21) / {(400 â€“ 609) / 420}

sec Î¸ / (tan Î¸ â€“ cosec Î¸) = (29 / 21) / {(-209 / 420)}

sec Î¸ / (tan Î¸ â€“ cosec Î¸) = (29 / 21) x (-420 / 209)

sec Î¸ / (tan Î¸ â€“ cosec Î¸) = – 580 / 209

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

(a) sin A cos A

(b) (cos A â€“ sin A) / (cos A + sin A)

Solution:

â–³ABC is a right-angled triangle.

By Pythagoras theorem,

We have,

AC2 = AB2 + BC2

AB2 = AC2 â€“ BC2

AB2 = 132 â€“ 122

AB2 = 169 â€“ 144

AB2 = 25

AB = âˆš25

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 â€“ sin A) / (cos A + sin A) = {(5/ 13) â€“ (12 / 13)} / {(5/ 13) + (12 / 13)}

(cos A â€“ sin A) / (cos A + sin A) = (-7 / 13) / (17 / 13)

(cos A â€“ sin A) / (cos A + sin A) = (-7 / 13) x (13 / 17)

(cos A â€“ sin A) / (cos A + sin A) = -7 / 17