OP Malhotra Class 9 Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(D)

Students can cross-reference their work with ICSE S Chand Maths Class 9 Solutions Chapter 19 Trigonometrical Ratios Ex 19(D) to ensure accuracy.

S Chand Class 9 ICSE Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(D)

Without using trigonometric tables, evaluate questions 1 and 2.

Question 1.
(i) \(\frac{\sin 16^{\circ}}{\cos 74^{\circ}}\)
(ii) \(\frac{\cos 25^{\circ}}{\sin 65^{\circ}}\)
(iii) \(\frac{\tan 38^{\circ}}{\cot 52^{\circ}}\)
(iv) \(\frac { sec 62° }{ cosec 28° }\)
Solution:
(i) \(\frac{\sin 16^{\circ}}{\cos 74^{\circ}}\) = \(\frac{\sin 16^{\circ}}{\cos \left(90^{\circ}-16^{\circ}\right)}\)
= \(\frac{\sin 16^{\circ}}{\sin 16^{\circ}}\) = 1 {∵ cos (90° – θ) = sin θ}

(ii) \(\frac{\cos 25^{\circ}}{\sin 65^{\circ}}\) = \(\frac{\cos 25^{\circ}}{\sin \left(90^{\circ}-25^{\circ}\right)}\)
= \(\frac{\cos 25^{\circ}}{\cos 25^{\circ}}\) = 1 {∵ sin (90° – θ) = cos θ}

(iii) \(\frac{\tan 38^{\circ}}{\cot 52^{\circ}}\) = \(\frac{\tan 38^{\circ}}{\cot \left(90^{\circ}-38^{\circ}\right)}\)
= \(\frac{\tan 38^{\circ}}{\tan 38^{\circ}}\) = 1 {∵ cot (90° – θ) = tan θ}

Question 2.
(i) sin² 67° + sin² 23°.

(ii) \(\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^2\) + \(\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^2\).

(iii) \(\frac{\cos ^2 20^{\circ}+\cos ^2 70^{\circ}}{\sin ^2 59^{\circ}+\sin ^2 31^{\circ}}\).

(iv) \(\frac{\cos 70^{\circ}}{\sin 20^{\circ}}\) + \(\frac{\cos 59^{\circ}}{\sin 31^{\circ}}\) – 8 sin² 30°

(v) 2\(\frac{\tan 53^{\circ}}{\cot 37^{\circ}}\) – \(\frac{\cot 80^{\circ}}{\tan 10^{\circ}}\)

(vi) sec 50° sin 40° + cos 40° cosec 50°.

(vii) \(\frac{\cos 80^{\circ}}{\sin 10^{\circ}}\) + cos 59° cosec 31°.

(viii) \(\frac{2 \sin 43^{\circ}}{\cos 47^{\circ}}\) – \(\frac{\cot 30^{\circ}}{\tan 60^{\circ}}\) – √2 sin 45°.

(ix) \(\frac{\cos ^2 20^{\circ}+\cos ^2 70^{\circ}}{\sin ^2 20^{\circ}+\sin ^2 70^{\circ}}\) + sin² 64° + cos 64° sin 26°.

(x) tan 35° tan 40° tan 45° tan 50° tan 55°.

(xi) sin² 10° + sin² 80° + \(\frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \sin 75^{\circ}}{\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos \left(90^{\circ}-\theta\right)}\)

(xii) \(\frac{\cos 75^{\circ}}{\sin 15^{\circ}}\) + \(\frac{\sin 12^{\circ}}{\cos 78^{\circ}}\) – \(\frac{\cos 18^{\circ}}{\sin 72^{\circ}}\)
Solution:
(i) sin² 67° + sin² 23°
= sin² 67° + sin² (90° – 67°)
= sin² 67° + cos² 67°
= 1 {∵ sin²θ + cos²θ = 1}

(ii) \(\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^2\) + \(\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^2\)
= \(\left(\frac{\sin 49^{\circ}}{\cos \left(90^{\circ}-49^{\circ}\right)}\right)^2\) + \(\left(\frac{\cos 41^{\circ}}{\sin \left(90^{\circ}-41^{\circ}\right)}\right)^2\)

= \(\left(\frac{\sin 49^{\circ}}{\sin 49^{\circ}}\right)^2\) + \(\left(\frac{\cos 41^{\circ}}{\cos 41^{\circ}}\right)^2\) = (1)² + (1)²
= 1 + 1 = 2

(iii) \(\frac{\cos ^2 20^{\circ}+\cos ^2 70^{\circ}}{\sin ^2 59^{\circ}+\sin ^2 31^{\circ}}\)
= \(\frac{\cos ^2\left(90^{\circ}-70^{\circ}\right)+\cos ^2 70^{\circ}}{\sin ^2 59^{\circ}+\sin ^2\left(90^{\circ}-59^{\circ}\right)^2}\)
= \(\frac{\sin ^2 70^{\circ}+\cos ^2 70^{\circ}}{\sin ^2 59^{\circ}+\cos ^2 59^{\circ}}\)

= \(\frac { 1 }{ 1 }\) = 1 {∵ sin²θ + cos²θ = 1}

(iv) \(\frac{\cos 70^{\circ}}{\sin 20^{\circ}}\) + \(\frac{\cos 59^{\circ}}{\sin 31^{\circ}}\) – 8 sin² 30°
= \(\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}\) + \(\frac{\cos \left(90^{\circ}-31^{\circ}\right)}{\sin 31^{\circ}}\) – 8 (sin 30°)²
= \(\frac{\sin 20^{\circ}}{\sin 20^{\circ}}\) + \(\frac{\sin 31^{\circ}}{\sin 31^{\circ}}\) – 8 \(\left(\frac{1}{2}\right)^2\)

= 1 + 1 – 8 × \(\frac { 1 }{ 4 }\) = 2 – 2 = 0

(v) 2\(\frac{\tan 53^{\circ}}{\cot 37^{\circ}}\) – \(\frac{\cot 80^{\circ}}{\tan 10^{\circ}}\)
= 2\(\frac{\tan \left(90^{\circ}-37^{\circ}\right)}{\cot 37^{\circ}}\) – \(\frac{\cot 80^{\circ}}{\tan \left(90^{\circ}-80^{\circ}\right)}\)
= 2\(\frac{\cot 37^{\circ}}{\cot 37^{\circ}}\) – \(\frac{\cot 80^{\circ}}{\cot 80^{\circ}}\) {∵ tan (90° – θ) = cot θ}
= 2 × 1 – 1 = 2 – 1 = 1

(vi) sec 50° sin 40° + cos 40° cosec 50°
= \(\frac{\sin 40^{\circ}}{\cos 50^{\circ}}\) + \(\frac{\cos 40^{\circ}}{\sin 50^{\circ}}\)

= \(\frac{\sin 40^{\circ}}{\cos \left(90^{\circ}-40^{\circ}\right)}\) + \(\frac{\cos 40^{\circ}}{\sin \left(90^{\circ}-40^{\circ}\right)}\)
= \(\frac{\sin 40^{\circ}}{\sin 40^{\circ}}\) + \(\frac{\cos 40^{\circ}}{\cos 40^{\circ}}\)

= 1 + 1 = 2

(vii) \(\frac{\cos 80^{\circ}}{\sin 10^{\circ}}\) + cos 59° cosec 31°
= \(\frac{\cos 80^{\circ}}{\sin \left(10^{\circ}\right)}\) + \(\frac{\cos 59^{\circ}}{\sin 31^{\circ}}\)

= \(\frac{\cos 80^{\circ}}{\sin \left(90^{\circ}-80^{\circ}\right)}\) + \(\frac{\cos 59^{\circ}}{\sin \left(90^{\circ}-59^{\circ}\right)}\)
= \(\frac{\cos 80^{\circ}}{\cos 80^{\circ}}\) + \(\frac{\cos 59^{\circ}}{\cos 59^{\circ}}\)
= 1 + 1 = 2

(x) tan 35° tan 40° tan 45° tan 50° tan 55°
= tan 35° tan 40° tan 45° tan (90° – 40°) tan (90° – 35°)
= tan 35° tan 40° tan 45° cot 40° cot 35°
= tan 35° × cot 35° tan 40° × cot 40° × cot 40° × tan 45°

= 1 × 1 × 1 = 1

(xii) \(\frac{\cos 75^{\circ}}{\sin 15^{\circ}}\) + \(\frac{\sin 12^{\circ}}{\cos 78^{\circ}}\) – \(\frac{\cos 18^{\circ}}{\sin 72^{\circ}}\)
= \(\frac{\cos 75^{\circ}}{\sin \left(90^{\circ}-75^{\circ}\right)}\) + \(\frac{\sin 12^{\circ}}{\cos \left(90^{\circ}-12^{\circ}\right)}\) = \(\frac{\cos \left(90^{\circ}-72^{\circ}\right)}{\sin 72^{\circ}}\)

= \(\frac{\cos 75^{\circ}}{\cos 75^{\circ}}\) + \(\frac{\sin 12^{\circ}}{\sin 12^{\circ}}\) – \(\frac{\sin 72^{\circ}}{\sin 72^{\circ}}\)
= 1 + 1 – 1 = 1

Question 3.
Express the following in terms of t-ratios of angles lying between 0° and 45°.
(i) cosec 69° + cot 69°
(ii) sin 85° + cosec 85°
Solution:
Using sin (90° – θ), cosec (90° – θ) and cot (90° – θ) we get,

(i) cosec 69° + cot 69°
= cosec (90° – 21°) + cot (90° – 21°)
= sec 21° + tan 21°

(ii) sin 85° + cosec 85°
= sin (90° – 5°) + cosec (90° – 5°)
= cos 5° + sec 5°

Question 4.
Prove that:

(ii) \(\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}\) + \(\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}\) = 2
(iii) sin θ sin (90° – θ) – cos θ cos (90° – θ) = 0
(iv) sec (90° – θ) cosec (90° – θ) = sec2θ cot θ

(vi) sin (60° – θ) = cos (30° – θ)
(vii) \(\frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)}\) + \(\frac{\cos \theta}{\cos \left(90^{\circ}-\theta\right)}\) = sec (90° – θ) cosec (90° – θ)
(viii) cos (81° + θ) = sin (9° – θ)
Solution:

(ii) \(\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}\) + \(\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}\) = 2
L.H.S. = \(\) + \(\)
= \(\frac{\cos \theta}{\cos \theta}\) + \(\frac{\sin \theta}{\sin \theta}\)

= 1 + 1 = 2 = R.H.S.

(iii) sin θ sin (90° – θ) – cos θ cos (90° – θ) = 0 L.H.S.
= sin θ sin (90° – θ) – cos θ cos (90° – θ)
= sin θ cos θ – cos θ sin θ

= 0 = R.H.S.

(iv) sec (90° – θ) cosec (90° – θ) = sec² θ cot θ
L.H.S.
= sec (90° – θ) cosec (90° – θ)
= cosec θ sec θ

R.H.S. = sec² θ cot θ = sec² θ × \(\frac{\cos \theta}{\sin \theta}\)
= \(\frac{\sec \theta \times \sec \theta \times \cos \theta}{\sin \theta}\) = \(\frac{\sec \theta \times 1}{\sin \theta}\)

= sec θ cosec θ = cosec θ sec θ
∴ L.H.S. = R.H.S.

(vi) sin (60° – θ) = cos (30° – θ)
L.H.S. = sin (60° – θ) = sin {90° – (30° + θ)
= cos (30° – θ) {sin (90° – θ) = cos θ)
= R.H.S.

(vii) \(\frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)}\) + \(\frac{\cos \theta}{\cos \left(90^{\circ}-\theta\right)}\) = sec (90° – θ)
cosec (90° – θ)
L.H.S. = \(\frac{\sin \theta}{\sin \left(90^{\circ}-\theta\right)}\) + \(\frac{\cos \theta}{\cos \left(90^{\circ}-\theta\right)}\)
= \(=\frac{\sin \theta}{\cos \theta}\) + \(\frac{\cos \theta}{\sin \theta}\) = \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}\) = \(\frac{1}{\sin \theta \cos \theta}\)

(viii) cos (81° + θ) = sin (9° – θ)
L.H.S. = cos (81° + θ) = cos {(90° – 9°) + θ}
= cos {90° – (9° – θ) = sin 9° – θ
= R.H.S. {cos (90° – θ) = sin θ}

Question 5.
(i) Find θ, if sin (θ + 36°) = cos θ, where θ + 36° is an acute angle.
(ii) Find the value of θ, if sin 5θ = cos 4θ, where 5θ and 4θ are acute angles.
(iii) If A is an acute angle, solve sin 3 A = cos 2 A.
Solution:
(i) sin (θ + 36°) = cos θ
⇒ sin (θ + 36°) = sin (90° – θ) { sin (90° – θ) = cos θ}
∴ θ + 36° = 90° – θ ⇒ θ + θ = 90° – 36°
= 2θ = 54° ⇒ θ = \(\frac{54^{\circ}}{2}\) = 27°
∴ θ = 27°

(ii) sin 5θ = cos 4 θ
⇒ sin 5θ = sin (90° – 4θ) {cos θ = sin (90° – θ)}
∴ 5θ = 90° – 4θ ⇒ 5θ + 4θ = 90°
⇒ 9θ = 90° ⇒ θ = \(\frac{90^{\circ}}{9}\) = 10°
∴ θ = 10°

(iii) sin 3A = cos 2A
sin 3A = sin (90° – 2A) {cos θ = sin (90° – θ)}
∴ 3A = 90° – 2A ⇒ 3A + 2A = 90°
⇒ 5A = 90° ⇒ A = \(\frac{90^{\circ}}{5}\) = 18°
∴ A = 18°