CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 25

Answer: (d) \( \sqrt{2} - 1 \)

Explanation:
1. cos(90° - θ) = sin θ, so sin θ + cos θ = √2 sin θ.
2. cos θ = (√2 - 1) sin θ.
3. cot θ = cos θ/sin θ = √2 - 1.

Teacher's Note:
1. Convert cos(90° - θ) into sin θ.
2. Divide by sin θ.

Answer: (b) \( \frac{1}{x} \)

Explanation:
1. (sec θ + tan θ)(sec θ - tan θ) = sec² θ - tan² θ = 1.
2. So sec θ - tan θ = 1/x.

Teacher's Note:
1. Use sec² θ - tan² θ = 1.
2. The two factors are reciprocals.

Answer: (a) \( a^2b^2 \)

Explanation:
1. b²x² + a²y² = b²a² sin² θ + a²b² cos² θ.
2. = a²b²(sin² θ + cos² θ) = a²b².

Teacher's Note:
1. Substitute x and y.
2. Use sin² θ + cos² θ = 1.

Answer: (a) true for all values of the variable

Explanation:
1. An identity holds for every value of the variable for which both sides are defined.
2. An equation true only for particular values is a conditional equation.

Teacher's Note:
1. Compare an identity with an equation like x + 1 = 3.
2. Think of sin² θ + cos² θ = 1.

Answer: (b) -1

Explanation:
1. 1/sin² θ = cosec² θ.
2. x = cot² θ - cosec² θ = -(cosec² θ - cot² θ) = -1.

Teacher's Note:
1. Use cosec² θ - cot² θ = 1.
2. Mind the sign.

Answer: (c) -1

Explanation:
1. sin⁶ + cos⁶ = 1 - 3 sin² cos² and sin⁴ + cos⁴ = 1 - 2 sin² cos².
2. Expression = 2(1 - 3 sin² cos²) - 3(1 - 2 sin² cos²) = 2 - 6 sin² cos² - 3 + 6 sin² cos² = -1.

Teacher's Note:
1. Write each sum in terms of sin² cos².
2. The sin² cos² terms cancel.

Answer: (a) \( \sec\theta + \tan\theta \)

Explanation:
1. Multiply numerator and denominator by (1 + sin θ): the fraction becomes (1 + sin θ)²/(1 - sin² θ) = (1 + sin θ)²/cos² θ.
2. Its square root is (1 + sin θ)/cos θ = sec θ + tan θ.

Teacher's Note:
1. Rationalise the denominator.
2. cos θ is positive for θ below 90°.

Answer: (b) \( 2\cos^2 A - 1 \)

Explanation:
1. cos⁴ A - sin⁴ A = (cos² A - sin² A)(cos² A + sin² A) = cos² A - sin² A.
2. = cos² A - (1 - cos² A) = 2 cos² A - 1.

Teacher's Note:
1. Factorise as a difference of squares.
2. Use sin² A = 1 - cos² A.

Answer: (b) 2

Explanation:
1. (1 + cot θ - cosec θ)(1 + tan θ + sec θ) = [(sin θ + cos θ - 1)/sin θ] × [(cos θ + sin θ + 1)/cos θ].
2. The numerator is (sin θ + cos θ)² - 1 = 2 sin θ cos θ.
3. So the product is 2 sin θ cos θ/(sin θ cos θ) = 2.

Teacher's Note:
1. Write everything in sin θ and cos θ.
2. Use (a - 1)(a + 1) = a² - 1 with a = sin θ + cos θ.

Answer: (b) 1

Explanation:
1. cosec θ - sin θ = cos² θ/sin θ and sec θ - cos θ = sin² θ/cos θ.
2. tan θ + cot θ = 1/(sin θ cos θ).
3. Product = (cos² θ/sin θ)(sin² θ/cos θ)(1/(sin θ cos θ)) = sin θ cos θ/(sin θ cos θ) = 1.

Teacher's Note:
1. Convert everything to sin θ and cos θ.
2. The factors cancel completely.

Answer: (b) 250

Explanation:
1. Let t = tan² α. Dividing by cos⁴ α: 15t² + 10 = 6(1 + t)², so 9t² - 12t + 4 = 0 and t = 2/3.
2. cosec² α = 1 + 1/t = 5/2, so 8 cosec⁶ α = 8 × 125/8 = 125.
3. sec² α = 1 + t = 5/3, so 27 sec⁶ α = 27 × 125/27 = 125. Sum = 250.

Teacher's Note:
1. Divide by cos⁴ α to get a quadratic in tan² α.
2. Find cosec² α and sec² α.

Answer: (d) \( b^2 - a^2 = 2ac \)

Explanation:
1. sin θ + cos θ = b/a and sin θ cos θ = c/a.
2. (sin θ + cos θ)² = 1 + 2 sin θ cos θ gives b²/a² = 1 + 2c/a.
3. b² = a² + 2ac, i.e. b² - a² = 2ac.

Teacher's Note:
1. Square the sum of the roots.
2. Use sin² + cos² = 1.

Answer: (c) 1

Explanation:
1. cos 47° = sin 43°, so cos² 47° - sin² 43° = 0.
2. sin 107° = sin 73°, so cos² 73° + sin² 107° = cos² 73° + sin² 73° = 1.

Teacher's Note:
1. Pair the terms with complementary and supplementary angles.
2. Use sin(180° - x) = sin x.

Answer: (d) \( -\frac{7\sqrt{5}}{5} \)

Explanation:
1. Product of roots = sin θ sec θ = tan θ = 1/2.
2. So sin θ = 1/√5, cos θ = 2/√5 and sec θ = √5/2.
3. Sum = 1/√5 + √5/2 = 7/(2√5) = -k/2, so k = -7/√5 = -7√5/5.

Teacher's Note:
1. The product of roots gives tan θ.
2. Find sin θ and sec θ from tan θ.

Answer: (b) \( \pm\sqrt{a^2 + b^2 - c^2} \)

Explanation:
1. (a cos θ - b sin θ)² + (a sin θ + b cos θ)² = a² + b².
2. So (a sin θ + b cos θ)² = a² + b² - c².

Teacher's Note:
1. Square both expressions and add.
2. Take the square root with both signs.

Answer: (c) 25

Explanation:
1. Square and add: (a cos θ + b sin θ)² + (a sin θ - b cos θ)² = a² + b².
2. a² + b² = 16 + 9 = 25.

Teacher's Note:
1. Square both equations and add them.
2. The cross terms cancel.

Answer: (d) 4

Explanation:
1. sin θ + sin³ θ = 1 - sin² θ = cos² θ, so sin θ(1 + sin² θ) = cos² θ.
2. Squaring: sin² θ(1 + sin² θ)² = cos⁴ θ. With c = cos² θ: (1 - c)(2 - c)² = c².
3. Expanding: (1 - c)(4 - 4c + c²) = 4 - 8c + 5c² - c³ = c², so c³ - 4c² + 8c = 4.
4. Hence cos⁶ θ - 4 cos⁴ θ + 8 cos² θ = 4.

Teacher's Note:
1. Square the relation sin θ(1 + sin² θ) = cos² θ.
2. Put c = cos² θ.

Answer: (c) \( \frac{1 - m}{1 + m}\cot\phi \)

Explanation:
1. Expand: cos θ cos φ - sin θ sin φ = m(cos θ cos φ + sin θ sin φ).
2. (1 - m) cos θ cos φ = (1 + m) sin θ sin φ.
3. tan θ = (1 - m)/(1 + m) × cos φ/sin φ = (1 - m)/(1 + m) × cot φ.

Teacher's Note:
1. Expand both cosines.
2. Collect cos θ cos φ and sin θ sin φ.

Answer: (b) 1

Explanation:
1. sin³ θ - cos³ θ = (sin θ - cos θ)(1 + sin θ cos θ).
2. Squaring sin θ - cos θ = 1: 1 - 2 sin θ cos θ = 1, so sin θ cos θ = 0.
3. The value is 1 × (1 + 0) = 1.

Teacher's Note:
1. Factorise the difference of cubes.
2. Square the given relation.

Answer: (c) 60°

Explanation:
1. Multiply by cos θ: √3 sin θ - 1 = cos θ, so √3 sin θ - cos θ = 1.
2. 2 sin(θ - 30°) = 1, so θ - 30° = 30°.
3. θ = 60°. Check: √3 × √3 - 2 = 1.

Teacher's Note:
1. Convert to sine and cosine.
2. Use the form R sin(θ - α).

Answer: (b) \( b^2 - a^2 \)

Explanation:
1. p² - q² = (a² - b²)cot² θ + (b² - a²)cosec² θ + (2ab - 2ab)cot θ cosec θ.
2. = (a² - b²)(cot² θ - cosec² θ) = -(a² - b²) = b² - a².

Teacher's Note:
1. Expand p² and q².
2. The cross terms cancel.

Answer: (b) \( \frac{5}{2} \)

Explanation:
1. (tan θ + cot θ)² = tan² θ + cot² θ + 2 = 17/4 + 2 = 25/4.
2. For acute θ both terms are positive, so tan θ + cot θ = 5/2.

Teacher's Note:
1. Square the required expression.
2. tan θ × cot θ = 1.

Answer: (a) \( \frac{-1 + \sqrt{5}}{2} \)

Explanation:
1. cos θ = sin θ/cos θ gives cos² θ = sin θ, i.e. 1 - sin² θ = sin θ.
2. sin² θ + sin θ - 1 = 0, so sin θ = (-1 ± √5)/2.
3. In the first quadrant sin θ is positive, so sin θ = (√5 - 1)/2.

Teacher's Note:
1. Multiply both sides by cos θ.
2. Reject the negative root.

Answer: (a) \( \sin\alpha \)

Explanation:
1. √(1 - sin² α) = cos α for acute α.
2. y = tan α cos α = sin α.

Teacher's Note:
1. Use 1 - sin² α = cos² α.
2. tan α cos α = sin α.

Answer: (a) \( \frac{\sqrt{3} + 1}{2\sqrt{3}} \)

Explanation:
1. sin 60° + cos 60° = (√3 + 1)/2 and tan 60° = √3.
2. The value = (√3 + 1)/(2√3).

Teacher's Note:
1. Substitute the standard values.
2. Divide by tan 60°.

Answer: (b) \( \sin^2 A \)

Explanation:
1. p = (1 + sin A)(1 - sin A)/cos A = cos A.
2. q = sec² A/cosec² A = tan² A.
3. p²q = cos² A tan² A = sin² A.

Teacher's Note:
1. Simplify p and q separately.
2. 1 + tan² A = sec² A and 1 + cot² A = cosec² A.

Answer: (b) \( \frac{\sin^2 A}{\cos A} \)

Explanation:
1. p = cos A and q = tan² A.
2. pq = cos A × sin² A/cos² A = sin² A/cos A.

Teacher's Note:
1. Use the values of p and q.
2. Simplify cos A × tan² A.

Answer: (b) \( \cos A - \tan A \)

Explanation:
1. p = cos A and √q = tan A for acute A.
2. p - √q = cos A - tan A.

Teacher's Note:
1. Take the positive square root of tan² A.
2. Substitute p.

Answer: (a) θ = 0° or θ = 30°

Explanation:
1. Replace cos² θ by 1 - sin² θ: sin² θ - (1/2) sin θ = 0.
2. sin θ(sin θ - 1/2) = 0, so sin θ = 0 or 1/2.
3. θ = 0° or 30°.

Teacher's Note:
1. Express everything in sin θ.
2. Factorise.