CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 16

Answer: (d) \( x^2q^2-y^2p^2=p^2q^2 \)

Explanation:
1. \( \sec\theta=\frac{x}{p} \) and \( \tan\theta=\frac{y}{q} \).
2. Use \( \sec^2\theta-\tan^2\theta=1 \): \( \frac{x^2}{p^2}-\frac{y^2}{q^2}=1 \).
3. Multiply by \( p^2q^2 \): \( x^2q^2-y^2p^2=p^2q^2 \).

Teacher's Note:
1. Use the identity sec² θ − tan² θ = 1.
2. Clear the denominators at the end.

Answer: (c) ≥ 2

Explanation:
1. sin x + cosec x = \( \sin x+\frac{1}{\sin x} \).
2. A positive number plus its reciprocal is at least 2.

Teacher's Note:
1. Here sin x is positive.
2. Equality at x = π/2.

Answer: (d) 0

Explanation:
1. sin 5θ = cos 4θ gives 5θ + 4θ = 90°, so θ = 10°.
2. 2 sin 30° − √3 tan 30° = 2 × 1/2 − √3 × 1/√3 = 1 − 1 = 0.

Teacher's Note:
1. If sin A = cos B for acute angles then A + B = 90°.
2. Then use standard values.

Answer: (c) \( \frac{65}{16} \)

Explanation:
1. sin A = 4/5, so cot A = 3/4.
2. 9 cot² A − 1 = 9 × 9/16 − 1 = 81/16 − 1 = 65/16.

Teacher's Note:
1. Find sin A from cos A first.
2. Then cot A = cos A / sin A.

Answer: (d) \( \frac{a+b}{a-b} \)

Explanation:
1. cosec θ = (a² + b²)/(a² − b²) and cot θ = \( \sqrt{\operatorname{cosec}^2\theta-1}=\frac{2ab}{a^2-b^2} \).
2. Sum = \( \frac{(a+b)^2}{a^2-b^2}=\frac{a+b}{a-b} \).

Teacher's Note:
1. Use cosec² θ − cot² θ = 1.
2. Factor the numerator as a perfect square.

Answer: (d) 7

Explanation:
1. Expand: sin² A + cosec² A + 2 + cos² A + sec² A + 2 = 1 + 4 + (1 + cot² A) + (1 + tan² A).
2. This equals 7 + tan² A + cot² A, so a = 7.

Teacher's Note:
1. Use sin² + cos² = 1, cosec² = 1 + cot², sec² = 1 + tan².
2. Collect constants.

Answer: (c) 45°

Explanation:
1. Add the fractions: \( \frac{2\cos\theta\operatorname{cosec}\theta}{\operatorname{cosec}^2\theta-1}=\frac{2\cos\theta\operatorname{cosec}\theta}{\cot^2\theta}=2\tan\theta \).
2. 2 tan θ = 2 gives tan θ = 1, so θ = 45°.

Teacher's Note:
1. Use cosec² θ − 1 = cot² θ.
2. Then cos θ cosec θ = cot θ.

Answer: (d) \( \frac{1}{\sin^2\theta-\cos^2\theta} \)

Explanation:
1. First term = \( \frac{\sin^2\theta}{\sin^2\theta-\cos^2\theta} \).
2. Since \( \sec^2\theta-\operatorname{cosec}^2\theta=\frac{\sin^2\theta-\cos^2\theta}{\sin^2\theta\cos^2\theta} \), the second term = \( \frac{\cos^2\theta}{\sin^2\theta-\cos^2\theta} \).
3. Sum = \( \frac{1}{\sin^2\theta-\cos^2\theta} \).

Teacher's Note:
1. Convert everything to sin and cos.
2. Combine over the common denominator.

Answer: (c) ab

Explanation:
1. \( x=\frac{a(1+\cos\theta)}{\sin\theta} \).
2. xy = \( \frac{ab(1-\cos^2\theta)}{\sin^2\theta}=ab \).

Teacher's Note:
1. Write cosec θ + cot θ as (1 + cos θ)/sin θ.
2. Use (1 − cos θ)(1 + cos θ) = sin² θ.

Answer: (b) \( \frac{225}{64} \)

Explanation:
1. Numerator = 2(1 − sin² θ) = 2cos² θ and denominator = 2(1 − cos² θ) = 2sin² θ.
2. The value is cot² θ = 225/64.

Teacher's Note:
1. Use (1 + x)(1 − x) = 1 − x².
2. Then use sin² + cos² = 1.

Answer: (b) 1

Explanation:
1. 1/sec θ = cos θ, and the largest value of cos θ is 1.

Teacher's Note:
1. Rewrite 1/sec θ as cos θ.
2. Cosine never exceeds 1.

Answer: (d) \( \frac{7}{25} \)

Explanation:
1. cosec² x − cot² x = 1, so cosec x + cot x = 3.
2. Adding: 2 cosec x = 10/3, so cosec x = 5/3, sin x = 3/5 and cos x = 4/5.
3. cos² x − sin² x = 16/25 − 9/25 = 7/25.

Teacher's Note:
1. Use the identity cosec² − cot² = 1.
2. Find sin x and cos x.

Answer: (d) \( \frac{\sqrt{3}}{2} \)

Explanation:
1. tan 30° = 1/√3, so the numerator is 2/√3 and the denominator is 1 + 1/3 = 4/3.
2. The value is \( \frac{2}{\sqrt{3}}\times\frac{3}{4}=\frac{\sqrt{3}}{2} \).

Teacher's Note:
1. The expression is the formula for sin 2θ with θ = 30°.
2. sin 60° = √3/2.