# Exercise 8.1

**1. In Δ 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 – 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 = √7 k

cos A = AB/AC = √7 k/4k = √7/4

tan A = BC/AB = 3k/√7 k = 3/√7

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

**Answer:**

#### Let ΔABC 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 θ = 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 θ =

#### = =

#### (iii)Tan θ =

#### = =

#### (iv) Cot θ =

#### = =

#### (v) Cosec θ =

#### = =

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

**Answer:**

#### cosA=cos

#### But

**7. If cot θ =7/8, evaluate :**

(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)

(ii) cot^{2}θ

**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 – sin^{2}A or not.

^{2}A)/(1+tan

^{2}A) = cos

^{2}A – 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/√3 find the value of:**

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

(ii) cos A cos C – sin A sin C

**Answer:**

#### Consider a triangle ABC in which

#### .

#### (i)

#### (ii)

**10. In Δ 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. ∴ 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

∴ 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 θ = 4/3 for some angle θ.

**Answer:**

#### i) False.

In ΔABC in which ∠B = 90º,

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 ΔABC in which ∠B = 90º,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 ∠A.

(v) False.

sin θ = Height/Hypotenuse

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

∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

# Exercise 8.2

**1. Evaluate the following :**

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan^{2}45° + cos^{2}30° – sin^{2}60°

(iii) cos 45°/(sec 30° + cosec 30°)

(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)

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

**Answer:**

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

#### (ii) 2 tan^{2}45° + cos^{2}30° – sin^{2}60°

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

#### (iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)

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

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

(i) 2tan 30°/1+tan^{2}30° =

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

(ii) 1-tan^{2}45°/1+tan^{2}45° =

(A) tan 90° (B) 1 (C) sin 45° (D) 0

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

(A) 0° (B) 30° (C) 45° (D) 60°

(iv) 2tan30°/1-tan^{2}30° =

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

**Answer:**

#### (i) 2tan 30°/1+tan^{2}30° =

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

#### =

#### (ii) 1-tan^{2}45°/1+tan^{2}45° =

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

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

#### (A) 0° (B) 30° (C) 45° (D) 60°

#### (iv) 2tan30°/1-tan^{2}30° =

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

**3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.**

**Answer:**

#### ……….(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 θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

**Answer:**

(i) False.

Let A = 30° and B = 60°, then

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

sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

(ii) True.

sin 0° = 0

sin 30° = 1/2

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Thus the value of sin θ increases as θ increases.

(iii) False.

cos 0° = 1

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

Thus the value of cos θ decreases as θ increases.

(iv) True.

cot A = cos A/sin A

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

# Exercise 8.3

**1. Evaluate :**

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

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

**Answer:**

#### (i) sin 18°/cos 72°

= sin (90° – 18°) /cos 72°

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

= tan (90° – 36°)/cot 64°

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

= cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

= cosec (90° – 59°) – sec 59°

= sec 59° – sec 59° = 0

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

**Answer:**

(i) tan 48° tan 23° tan 42° tan 67°

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

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

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

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

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

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

**Answer:**

tan 2A = cot (A- 18°)

⇒ cot (90° – 2A) = cot (A -18°)

Equating angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

⇒ A = 36°

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

**Answer:**

tan A = cot B

⇒ tan A = tan (90° – B)

⇒ A = 90° – B

⇒ A + B = 90°

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

**Answer:**

sec 4A = cosec (A – 20°)

⇒ cosec (90° – 4A) = cosec (A – 20°)

Equating angles,

90° – 4A= A- 20°

⇒ 110° = 5A

⇒ A = 22°

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

**Answer:**

sin 67° + cos 75°

= sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°

# 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 ∠A in terms of sec A.**

**Answer:**

**3. Evaluate :**

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

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

^{2}63° + sin

^{2}27°)/(cos

^{2}17° + cos

^{2}73°)

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

**Answer:**

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

^{2}63° + sin

^{2}27°)/(cos

^{2}17° + cos

^{2}73°)

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

#### 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 θ + sec θ) (1 + cot θ – cosec θ)**

(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 θ + sec θ) (1 + cot θ – cosec θ)**

(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 θ – cot θ)^{2} = (1-cos θ)/(1+cos θ)

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

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

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

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

[Hint : Simplify LHS and RHS separately]

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

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

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

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

(ix) (cosec A – sin A)(sec A – 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 θ – cot θ)^{2} = (1-cos θ)/(1+cos θ)

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

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

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

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

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

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

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

LHS = 1 + sin A/(1 – sin A) …..(1)

Multiplying and dividing by (1 + sin A)

⇒ (1 + sin A)(1 + sin A/1 – sin A)(1 + sin A)

= (1 + sin A)²/(1 – sin² A) [a² – b² = (a – b)(a + b)]

= (1 + sinA)/1 – sin² A

= 1 + sin A/cos² A

= 1 + sin A/cos A

= 1/cos A + sin A/cos A

= sec A + tan A

= R.H.S

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

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

(ix) (cosec A – sin A)(sec A – 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