# Exercise 8.1

**1. In Î” ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :**

(i) sin A, cos A

(ii) sin C, cos C

**Answer:**

#### Let us draw a right angled triangle ABC, right angled at B.

#### Using Pythagoras theorem,

#### (i)

#### (ii)

#### ,

**2. In adjoining figure, find tan P â€“ cot R.**

**Answer:**

**3. If sin A =3/4, calculate cos A and tan A.**

**Answer:**

#### Given: A triangle ABC in which

#### B =90

#### We know that sin A = BC/AC = 3/4

Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem we get,

AC^{2} = AB^{2} + BC^{2}

(4k)^{2} = AB^{2} + (3k)^{2}

16k^{2} – 9k^{2 }= AB^{2}

AB^{2} = 7k^{2}

AB = âˆš7 k

cos A = AB/AC = âˆš7 k/4k = âˆš7/4

tan A = BC/AB = 3k/âˆš7 k = 3/âˆš7

**4. Given 15 cot A = 8, find sin A and sec A.**

**Answer:**

#### Let Î”ABC be a right-angled triangle, right-angled at B.

We know that cot A = AB/BC = 8/15 (Given)

Let AB be 8k and BC will be 15k where k is a positive real number.

By Pythagoras theorem we get,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (8k)^{2} + (15k)^{2}

AC^{2 }= 64k^{2 }+ 225k^{2}

AC^{2} = 289k^{2}

AC = 17 k

sin A = BC/AC = 15k/17k = 15/17

sec A = AC/AB = 17k/8 k = 17/8

**5. Given sec Î¸ = 13/12, calculate all other trigonometric ratios.**

**Answer:**

#### Consider a triangle ABC in which

#### Let AB = 12k and AC = 13k

#### Then, using Pythagoras theorem,

#### (AC)2 = (AB)2 + (BC)2

#### (13 *k*)2 = (12 *k*)2 + (BC)2

#### 169 *k*2 = 144 *k*2 + BC2

#### 25 *k*2 = BC2

#### BC = 5 *k*

#### (I) Sin =

#### = =

#### (ii) Cos Î¸ =

#### = =

#### (iii)Tan Î¸ =

#### = =

#### (iv) Cot Î¸ =

#### = =

#### (v) Cosec Î¸ =

#### = =

**6. If âˆ A and âˆ B are acute angles such that cos A = cos B, then show that âˆ A = âˆ B.**

**Answer:**

#### cosA=cos

#### But

**7. If cot Î¸ =7/8, evaluate :**

(i)(1+sin Î¸ )(1-sin Î¸)/(1+cos Î¸)(1-cos Î¸)

(ii) cot^{2}Î¸

**Answer:**

#### Consider a triangle ABC

#### (ii)

**8. If 3cot A = 4/3 , check whether (1-tan**^{2}A)/(1+tan^{2}A) = cos^{2}A â€“ sin^{2}A or not.

^{2}A)/(1+tan

^{2}A) = cos

^{2}A â€“ sin

^{2}A or not.

**Answer:**

#### Consider a triangle ABC AB=4cm, BC= 3cm

#### .

#### And

**9. In triangle ABC, right-angled at B, if tan A =1/âˆš3 find the value of:**

#### (i) sin A cos C + cos A sin C

(ii) cos A cos C â€“ sin A sin C

**Answer:**

#### Consider a triangle ABC in which

#### .

#### (i)

#### (ii)

**10. In Î” PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.**

**Answer:**

Given that, PR + QR = 25 , PQ = 5

Let PR be x. âˆ´ QR = 25 – x

By Pythagoras theorem ,

PR2 = PQ^{2} + QR^{2}

x^{2 }= (5)2 + (25 – x)^{2}

x^{2} = 25 + 625 + x^{2} – 50x

50x = 650

x = 13

âˆ´ PR = 13 cm

QR = (25 – 13) cm = 12 cm

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

**11. State whether the following are true or false. Justify your answer.**

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin Î¸ = 4/3 for some angle Î¸.

**Answer:**

#### i) False.

In Î”ABC in which âˆ B = 90Âº,

AB = 3, BC = 4 and AC = 5

Value of tan A = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as

it will follow the Pythagoras theorem.

AC^{2 }= AB^{2} + BC^{2}

5^{2} = 3^{2 }+ 4^{2}

25 = 9 + 16

25 = 25

(ii) True.

Let a Î”ABC in which âˆ B = 90Âº,AC be 12k and AB be 5k, where k is a positive real number.

By Pythagoras theorem we get,

AC^{2} = AB^{2} + BC^{2}

(12k)^{2} = (5k)^{2} + BC^{2}

BC^{2 }+ 25k^{2} = 144k^{2}

BC^{2} = 119k^{2}

Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) False.

Abbreviation used for cosecant of angle A is cosec A.cos A is the abbreviation used for cosine of angle A.

(iv) False.

cot A is not the product of cot and A. It is the cotangent of âˆ A.

(v) False.

sin Î¸ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

âˆ´ sin Î¸ will always less than 1 and it can never be 4/3 for any value of Î¸.

# Exercise 8.2

**1. Evaluate the following :**

(i) sin 60Â° cos 30Â° + sin 30Â° cos 60Â°

(ii) 2 tan^{2}45Â° + cos^{2}30Â° â€“ sin^{2}60Â°

(iii) cos 45Â°/(sec 30Â° + cosec 30Â°)

(iv) (sin 30Â° + tan 45Â° â€“ cosec 60Â°)/(sec 30Â° + cos 60Â° + cot 45Â°)

(v) (5cos^{2}60Â° + 4sec^{2}30Â° – tan^{2}45Â°)/(sin^{2}30Â° + cos^{2}30Â°)

**Answer:**

#### (i) sin 60Â° cos 30Â° + sin 30Â° cos 60Â°

#### (ii) 2 tan^{2}45Â° + cos^{2}30Â° â€“ sin^{2}60Â°

#### (iii) cos 45Â°/(sec 30Â° + cosec 30Â°)

#### (iv) (sin 30Â° + tan 45Â° â€“ cosec 60Â°)/(sec 30Â° + cos 60Â° + cot 45Â°)

#### (v) (5cos^{2}60Â° + 4sec^{2}30Â° – tan^{2}45Â°)/(sin^{2}30Â° + cos^{2}30Â°)

**2. Choose the correct option and justify your choice :**

(i) 2tan 30Â°/1+tan^{2}30Â° =

(A) sin 60Â° (B) cos 60Â° (C) tan 60Â° (D) sin 30Â°

(ii) 1-tan^{2}45Â°/1+tan^{2}45Â° =

(A) tan 90Â° (B) 1 (C) sin 45Â° (D) 0

(iii) sin ^{2}A = 2 sin A is true when A =

(A) 0Â° (B) 30Â° (C) 45Â° (D) 60Â°

(iv) 2tan30Â°/1-tan^{2}30Â° =

(A) cos 60Â° (B) sin 60Â° (C) tan 60Â° (D) sin 30Â°

**Answer:**

#### (i) 2tan 30Â°/1+tan^{2}30Â° =

(A) sin 60Â° (B) cos 60Â° (C) tan 60Â° (D) sin 30Â°

#### =

#### (ii) 1-tan^{2}45Â°/1+tan^{2}45Â° =

#### (A) tan 90Â° (B) 1 (C) sin 45Â° (D) 0

#### (iii) sin ^{2}A = 2 sin A is true when A =

#### (A) 0Â° (B) 30Â° (C) 45Â° (D) 60Â°

#### (iv) 2tan30Â°/1-tan^{2}30Â° =

#### (A) cos 60Â° (B) sin 60Â° (C) tan 60Â° (D) sin 30Â°

**3. If tan (A + B) = âˆš3 and tan (A â€“ B) = 1/âˆš3; 0Â° < A + B â‰¤ 90Â°; A > B, find A and B.**

**Answer:**

#### â€¦â€¦â€¦.(i)

#### **4. State whether the following are true or false. Justify your answer.**

#### (i) sin (A + B) = sin A + sin B.

(ii) The value of sin Î¸ increases as Î¸ increases.

(iii) The value of cos Î¸ increases as Î¸ increases.

(iv) sin Î¸ = cos Î¸ for all values of Î¸.

(v) cot A is not defined for A = 0Â°.

**Answer:**

(i) False.

Let A = 30Â° and B = 60Â°, then

sin (A + B) = sin (30Â° + 60Â°) = sin 90Â° = 1 and,

sin A + sin B = sin 30Â° + sin 60Â°

= 1/2 + âˆš3/2 = 1+âˆš3/2

(ii) True.

sin 0Â° = 0

sin 30Â° = 1/2

sin 45Â° = 1/âˆš2

sin 60Â° = âˆš3/2

sin 90Â° = 1

Thus the value of sin Î¸ increases as Î¸ increases.

(iii) False.

cos 0Â° = 1

cos 30Â° = âˆš3/2

cos 45Â° = 1/âˆš2

cos 60Â° = 1/2

cos 90Â° = 0

Thus the value of cos Î¸ decreases as Î¸ increases.

(iv) True.

cot A = cos A/sin A

cot 0Â° = cos 0Â°/sin 0Â° = 1/0 = undefined.

# Exercise 8.3

**1. Evaluate :**

(i) sin 18Â°/cos 72Â° (ii) tan 26Â°/cot 64Â° (iii) cos 48Â° â€“ sin 42Â° (iv) cosec 31Â° â€“ sec 59Â°

(i) sin 18Â°/cos 72Â° (ii) tan 26Â°/cot 64Â° (iii) cos 48Â° â€“ sin 42Â° (iv) cosec 31Â° â€“ sec 59Â°

**Answer:**

#### (i) sin 18Â°/cos 72Â°

= sin (90Â° – 18Â°) /cos 72Â°

= cos 72Â° /cos 72Â° = 1

(ii) tan 26Â°/cot 64Â°

= tan (90Â° – 36Â°)/cot 64Â°

= cot 64Â°/cot 64Â° = 1

(iii) cos 48Â° – sin 42Â°

= cos (90Â° – 42Â°) – sin 42Â°

= sin 42Â° – sin 42Â° = 0

(iv) cosec 31Â° – sec 59Â°

= cosec (90Â° – 59Â°) – sec 59Â°

= sec 59Â° – sec 59Â° = 0

**2. Show that :
(i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1
(ii) cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â° = 0**

**Answer:**

(i) tan 48Â° tan 23Â° tan 42Â° tan 67Â°

= tan (90Â° – 42Â°) tan (90Â° – 67Â°) tan 42Â° tan 67Â°

= cot 42Â° cot 67Â° tan 42Â° tan 67Â°

= (cot 42Â° tan 42Â°) (cot 67Â° tan 67Â°) = 1Ã—1 = 1

(ii) cos 38Â° cos 52Â° – sin 38Â° sin 52Â°

= cos (90Â° – 52Â°) cos (90Â°-38Â°) – sin 38Â° sin 52Â°

= sin 52Â° sin 38Â° – sin 38Â° sin 52Â° = 0

**3. If tan 2A = cot (A â€“ 18Â°), where 2A is an acute angle, find the value of A.**

**Answer:**

tan 2A = cot (A- 18Â°)

â‡’ cot (90Â° – 2A) = cot (A -18Â°)

Equating angles,

â‡’ 90Â° – 2A = A- 18Â° â‡’ 108Â° = 3A

â‡’ A = 36Â°

**4. If tan A = cot B, prove that A + B = 90Â°.**

**Answer:**

tan A = cot B

â‡’ tan A = tan (90Â° – B)

â‡’ A = 90Â° – B

â‡’ A + B = 90Â°

**5. If sec 4A = cosec (A â€“ 20Â°), where 4A is an acute angle, find the value of A.**

**Answer:**

sec 4A = cosec (A – 20Â°)

â‡’ cosec (90Â° – 4A) = cosec (A – 20Â°)

Equating angles,

90Â° – 4A= A- 20Â°

â‡’ 110Â° = 5A

â‡’ A = 22Â°

**6. If A, B and C are interior angles of a triangle ABC, then show that**

sin (B+C/2) = cos A/2

sin (B+C/2) = cos A/2

**Answer:**

**7. Express sin 67Â° + cos 75Â° in terms of trigonometric ratios of angles between 0Â° and 45Â°.**

**Answer:**

sin 67Â° + cos 75Â°

= sin (90Â° – 23Â°) + cos (90Â° – 15Â°)

= cos 23Â° + sin 15Â°

# EXERCISE 8.4

**1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.**

**2. Write all the other trigonometric ratios of âˆ A in terms of sec A.**

**Answer:**

**3. Evaluate :**

**(i) (sin**^{2}63Â° + sin^{2}27Â°)/(cos^{2}17Â° + cos^{2}73Â°)

(ii) sin 25Â° cos 65Â° + cos 25Â° sin 65Â°

^{2}63Â° + sin

^{2}27Â°)/(cos

^{2}17Â° + cos

^{2}73Â°)

(ii) sin 25Â° cos 65Â° + cos 25Â° sin 65Â°

**Answer:**

#### **(i) (sin**^{2}63Â° + sin^{2}27Â°)/(cos^{2}17Â° + cos^{2}73Â°)

^{2}63Â° + sin

^{2}27Â°)/(cos

^{2}17Â° + cos

^{2}73Â°)

**(ii) sin 25Â° cos 65Â° + cos 25Â° sin 65Â°**

#### Choose the correct option. Justify your choice.

**4. (i) 9 sec**^{2}A – 9 tan^{2}A =

(A) 1 (B) 9 (C) 8 (D) 0

**(ii) (1 + tan Î¸ + sec Î¸) (1 + cot Î¸ – cosec Î¸)**

(A) 0 (B) 1 (C) 2 (D) – 1

**(iii) (secA + tanA) (1 – sinA) =**

(A) secA (B) sinA (C) cosecA (D) cosA

**(iv) 1+tan**^{2}A/1+cot^{2}A =

(A) sec^{2}A (B) -1 (C) cot^{2}A (D) tan^{2}A

^{2}A – 9 tan

^{2}A =

^{2}A/1+cot

^{2}A =

**Answer:**

#### (i) ** (i) 9 sec**^{2}A – 9 tan^{2}A =

^{2}A – 9 tan

^{2}A =

#### (A) 1 (B) 9 (C) 8 (D) 0

**(ii) (1 + tan Î¸ + sec Î¸) (1 + cot Î¸ – cosec Î¸)**

(A) 0 (B) 1 (C) 2 (D) – 1

**(iii) (secA + tanA) (1 – sinA) =**

(A) secA (B) sinA (C) cosecA (D) cosA

**(iv) 1+tan**^{2}A/1+cot^{2}A =

(A) sec^{2}A (B) -1 (C) cot^{2}A (D) tan^{2}A

^{2}A/1+cot

^{2}A =

**5. Prove the following identities, where the angles involved are acute angles for which the**

**expressions are defined.**

(i) (cosec Î¸ – cot Î¸)^{2} = (1-cos Î¸)/(1+cos Î¸)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(iii) tan Î¸/(1-cot Î¸) + cot Î¸/(1-tan Î¸) = 1 + sec Î¸ cosec Î¸

[Hint : Write the expression in terms of sin Î¸ and cos Î¸]

(iv) (1 + sec A)/sec A = sin^{2}A/(1-cos A)

[Hint : Simplify LHS and RHS separately]

(v) (cos Aâ€“sin A+1)/(cos A+sin Aâ€“1) = cosec A + cot A,using the identity cosec^{2}A = 1+cot^{2}A.

#### (vi) âˆš1 + sin A/1 – sin A = sec A+ tan A

#### (vii) (sin Î¸ – 2sin^{3}Î¸)/(2cos^{3}Î¸-cos Î¸) = tan Î¸

#### (viii) (sin A + cosec A)^{2} + (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

(ix) (cosec A â€“ sin A)(sec A â€“ cos A) = 1/(tan A+cotA)

[Hint : Simplify LHS and RHS separately]

(x) (1+tan^{2}A/1+cot^{2}A) = (1-tan A/1-cot A)^{2} = tan^{2}A

**Answer:**

#### (i) (cosec Î¸ – cot Î¸)^{2} = (1-cos Î¸)/(1+cos Î¸)

#### (ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

#### (iii) tan Î¸/(1-cot Î¸) + cot Î¸/(1-tan Î¸) = 1 + sec Î¸ cosec Î¸

#### [Hint : Write the expression in terms of sin Î¸ and cos Î¸]

#### (iv) (1 + sec A)/sec A = sin^{2}A/(1-cos A)

#### [Hint : Simplify LHS and RHS separately]

#### (v) (cos Aâ€“sin A+1)/(cos A+sin Aâ€“1) = cosec A + cot A,using the identity cosec^{2}A = 1+cot^{2}A

## (vi) âˆš1 + sin A/1 – sin A = sec A+ tan A

LHS = 1 + sin A/(1 – sin A) …..(1)

Multiplying and dividing by (1 + sin A)

â‡’ (1 + sin A)(1 + sin A/1 – sin A)(1 + sin A)

= (1 + sin A)Â²/(1 – sinÂ² A) [aÂ² – bÂ² = (a – b)(a + b)]

= (1 + sinA)/1 – sinÂ² A

= 1 + sin A/cosÂ² A

= 1 + sin A/cos A

= 1/cos A + sin A/cos A

= sec A + tan A

= R.H.S

(vii) (sin Î¸ – 2sin^{3}Î¸)/(2cos^{3}Î¸-cos Î¸) = tan Î¸

(viii) (sin A + cosec A)^{2} + (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

(ix) (cosec A â€“ sin A)(sec A â€“ cos A) = 1/(tan A+cotA)

[Hint : Simplify LHS and RHS separately]

(x) (1+tan^{2}A/1+cot^{2}A) = (1-tan A/1-cot A)^{2} = tan^{2}A