6. Given 15 cot A = 8, then find the values of sin A and sec A [CBSE 2020] (2 marks)

Solution:

15 cot A = 8 ⇒ cot A = 8/15

We know: cot A = adjacent/opposite = 8/15

So, adjacent = 8, opposite = 15

Hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17

sin A = opposite/hypotenuse = 15/17

sec A = hypotenuse/adjacent = 17/8

Answer: sin A = 15/17, sec A = 17/8

4. If sin θ = cos θ, then the value of tan 2θ + cot² θ is [CBSE 2020]

(a) 2

(b) 4

(c) 1

(d) 10/3

Solution:

sin θ = cos θ ⇒ θ = 45°

tan 2θ = tan 90° = undefined

cot² θ = cot² 45° = 1² = 1

This question seems to have an issue as tan 90° is undefined.

Let’s re-examine: If sin θ = cos θ, then θ = 45°

tan 2θ + cot² θ = tan 90° + cot² 45° = undefined + 1 = undefined

This doesn’t match any option, suggesting there might be a typo in the question.

33. The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ – b cos θ) is [CBSE 2020]

(a) a² + b²

(b) a² – b²

(c) √(a² + b²)

(d) √(a² – b²)

Solution:

Distance = √[(a cos θ + b sin θ – 0)² + (0 – (a sin θ – b cos θ))²]

= √[(a cos θ + b sin θ)² + (-a sin θ + b cos θ)²]

= √[a² cos²θ + 2ab sinθ cosθ + b² sin²θ + a² sin²θ – 2ab sinθ cosθ + b² cos²θ]

= √[a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ)]

= √[a² + b²]

Answer: (c) √(a² + b²)

13. Evaluate: 2 sec 30° × tan 60° [CBSE 2020]

Solution:

sec 30° = 2/√3, tan 60° = √3

2 × (2/√3) × √3 = 2 × 2 = 4

Answer: 4

14. Write the value of sin² 30° + cos² 60° [CBSE 2020]

Solution:

sin 30° = 1/2, cos 60° = 1/2

sin² 30° + cos² 60° = (1/2)² + (1/2)² = 1/4 + 1/4 = 1/2

Answer: 1/2

15. Evaluate: \(\frac{2\tan 45^\circ \times \cos 60^\circ}{\sin 30^\circ}\) [CBSE 2020]

Solution:

tan 45° = 1, cos 60° = 1/2, sin 30° = 1/2

\(\frac{2 \times 1 \times \frac{1}{2}}{\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2\)

Answer: 2

36. 5 tan²θ – 5 sec²θ = ______ [CBSE 2020]

Solution:

5 tan²θ – 5 sec²θ = 5(tan²θ – sec²θ)

We know: 1 + tan²θ = sec²θ ⇒ tan²θ – sec²θ = -1

So, 5 × (-1) = -5

Answer: -5

37. Simplest form of (1 – cos² A)(1 + cot² A) is ______ [CBSE 2020]

Solution:

(1 – cos² A)(1 + cot² A) = sin² A × csc² A

= sin² A × (1/sin² A) = 1

Answer: 1

38. Simplest form of \(\frac{1 + \tan^2 A}{1 + \cot^2 A}\) is ______ [CBSE 2020]

Solution:

\(\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A\)

Answer: tan² A

39. The value of \(\left(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\right)\) = ______ [CBSE 2020]

Solution:

\(\sin^2 \theta + \frac{1}{1 + \tan^2 \theta} = \sin^2 \theta + \frac{1}{\sec^2 \theta} = \sin^2 \theta + \cos^2 \theta = 1\)

Answer: 1

40. The value of (1 + tan²θ)(1 – sinθ)(1 + sinθ) [CBSE 2020]

Solution:

(1 + tan²θ)(1 – sinθ)(1 + sinθ) = sec²θ × (1 – sin²θ)

= sec²θ × cos²θ = (1/cos²θ) × cos²θ = 1

Answer: 1

5. If tan A = 3/4, find the value of 1/sin A + 1/cos A [CBSE 2020]

Solution:

tan A = 3/4 ⇒ opposite = 3, adjacent = 4

Hypotenuse = √(3² + 4²) = √(9 + 16) = √25 = 5

sin A = 3/5, cos A = 4/5

1/sin A + 1/cos A = 5/3 + 5/4 = (20 + 15)/12 = 35/12

Answer: 35/12

12. If √3 sin θ – cos θ = 0 and 0° < θ < 90°, find the value of θ [CBSE 2020]

Solution:

√3 sin θ – cos θ = 0 ⇒ √3 sin θ = cos θ

tan θ = 1/√3 ⇒ θ = 30°

Answer: 30°

9. sin A + cos B = 1, A = 30° and B is an acute angle, then find the value of B [CBSE 2020]

Solution:

sin 30° + cos B = 1 ⇒ 1/2 + cos B = 1

cos B = 1 – 1/2 = 1/2 ⇒ B = 60°

Answer: 60°

55. Show that \(\frac{1 + \tan A}{2 \sin A} + \frac{1 + \cot A}{2 \cos A} = \csc A + \sec A\) [CBSE 2020]

Solution:

LHS = \(\frac{1 + \tan A}{2 \sin A} + \frac{1 + \cot A}{2 \cos A}\)

= \(\frac{1 + \frac{\sin A}{\cos A}}{2 \sin A} + \frac{1 + \frac{\cos A}{\sin A}}{2 \cos A}\)

= \(\frac{\frac{\cos A + \sin A}{\cos A}}{2 \sin A} + \frac{\frac{\sin A + \cos A}{\sin A}}{2 \cos A}\)

= \(\frac{\cos A + \sin A}{2 \sin A \cos A} + \frac{\sin A + \cos A}{2 \sin A \cos A}\)

= \(\frac{2(\sin A + \cos A)}{2 \sin A \cos A} = \frac{\sin A + \cos A}{\sin A \cos A}\)

= \(\frac{1}{\sin A} + \frac{1}{\cos A} = \csc A + \sec A\) = RHS

Hence proved.

56. Prove that: \(\frac{2 \cos^3 \theta – \cos \theta}{\sin \theta – 2 \sin^3 \theta} = \cot \theta\) [CBSE 2020]

Solution:

LHS = \(\frac{2 \cos^3 \theta – \cos \theta}{\sin \theta – 2 \sin^3 \theta} = \frac{\cos \theta (2 \cos^2 \theta – 1)}{\sin \theta (1 – 2 \sin^2 \theta)}\)

Using identities: 2 cos²θ – 1 = cos 2θ, 1 – 2 sin²θ = cos 2θ

= \(\frac{\cos \theta \cdot \cos 2\theta}{\sin \theta \cdot \cos 2\theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta\) = RHS

Hence proved.

57. Prove that: \((\sin^4 \theta – \cos^4 \theta + 1) \csc^2 \theta = 2\) [CBSE 2020]

Solution:

LHS = \((\sin^4 \theta – \cos^4 \theta + 1) \csc^2 \theta\)

= \(((\sin^2 \theta – \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) + 1) \csc^2 \theta\)

= \(((\sin^2 \theta – \cos^2 \theta) \cdot 1 + 1) \csc^2 \theta\)

= \((\sin^2 \theta – \cos^2 \theta + 1) \csc^2 \theta\)

= \((\sin^2 \theta + (1 – \cos^2 \theta)) \csc^2 \theta = (\sin^2 \theta + \sin^2 \theta) \csc^2 \theta\)

= \(2 \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 2\) = RHS

Hence proved.

58. Prove that: \(\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A\) [CBSE 2020]

Solution:

LHS = \(\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}\)

= \(\frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A\) = RHS

Hence proved.

59. If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1 [CBSE 2020]

Solution:

Given: sin θ + cos θ = √3

Squaring both sides: (sin θ + cos θ)² = 3

sin²θ + cos²θ + 2 sin θ cos θ = 3

1 + 2 sin θ cos θ = 3 ⇒ sin θ cos θ = 1

Now, tan θ + cot θ = \(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{1} = 1\)

Hence proved.