# Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios

Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios provide solutions in a stepwise manner as per studentsâ€™ intelligence quotient. The main aim of developing these solutions is to make the learning process smooth and hassle-free among students. Those who aspire to excel in solving difficult problems easily, can make use of Frank Solutions. Here, students can find the Frank Solutions Class 9 Maths Chapter 26 Trigonometrical Ratios PDF, from the links provided below.

Chapter 26 Trigonometrical Ratios has problems pertaining to the measurement of the sides and angles of a triangle. The branch of Mathematics, which deals with the study of sides and angles of a right-angled triangle is known as trigonometry. It is most essential to understand this chapter in-depth, as it plays a vital role in further classes also.

## 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

â–³ADB is a right angled triangle

Therefore,

AB2 = 82 + 62

AB2 = 64 + 36

AB2 = 100

We get,

AB = 10 cm

â–³ADC is a right angled triangle

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 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