IMO Maths Olympiad Class 10 Chapter 08 Introduction to Trigonometry MCQs with Answers

Answer: (c) 0

Explanation:
1. Taking reciprocals, the expression is \( u^3 + v^3 \) with \( u = \dfrac{1-2\sin A}{\sqrt{3}+2\cos A} \) and \( v = \dfrac{\sqrt{3}-2\cos A}{1+2\sin A} \).
2. \( \dfrac{u}{v} = \dfrac{(1-2\sin A)(1+2\sin A)}{(\sqrt{3}+2\cos A)(\sqrt{3}-2\cos A)} = \dfrac{1-4\sin^2 A}{3-4\cos^2 A} = \dfrac{1-4\sin^2 A}{4\sin^2 A-1} = -1 \).
3. So v = −u, and u³ + v³ = 0.

Teacher's Note:
1. Look for a relation between the two brackets instead of expanding the cubes.
2. 3 − 4cos²A = 4sin²A − 1 because sin²A + cos²A = 1.

Answer: (b) cot² B

Explanation:
1. Since B = 90° − A: tan A tan B = 1, tan A cot B = tan²A, sin B = cos A and sec B = cosec A.
2. So the first fraction is \( \dfrac{1+\tan^2 A}{\sin A \cdot \dfrac{1}{\sin A}} = \sec^2 A \) and the second is \( \dfrac{\cos^2 A}{\cos^2 A} = 1 \).
3. The value is sec²A − 1 = tan²A = cot²B (because cot B = tan A).

Teacher's Note:
1. Convert everything to angle A using B = 90° − A.
2. tan A and cot B are equal for complementary angles.

Answer: (d) m² + n²

Explanation:
1. Square both: m² = a²cos²θ + b²sin²θ + 2ab sin θ cos θ and n² = a²sin²θ + b²cos²θ − 2ab sin θ cos θ.
2. Add: m² + n² = a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ).
3. So m² + n² = a² + b².

Teacher's Note:
1. Squaring and adding is the standard way to eliminate θ.
2. The cross terms 2ab sin θ cos θ cancel.

Answer: (c) 60°, 30°, 0°

Explanation:
1. sin(A + B + C) = 1 gives A + B + C = 90°; tan(A − B) = 1/√3 gives A − B = 30°; sec(A + C) = 2 gives A + C = 60°.
2. From the first and third, B = 30°. Then A = 30° + 30° = 60° and C = 60° − 60° = 0°.
3. So A, B, C = 60°, 30°, 0°.

Teacher's Note:
1. Turn each trigonometric equation into a linear equation in the angles.
2. Only option (c) satisfies all three equations (the book accepts 0° as a limiting acute value).

Answer: (a) \( \dfrac{1}{\sin^2 A-\sin^4 A} \)

Explanation:
1. 1 + tan²A = sec²A and 1 + cot²A = cosec²A.
2. Sum = \( \dfrac{1}{\cos^2 A} + \dfrac{1}{\sin^2 A} = \dfrac{\sin^2 A + \cos^2 A}{\sin^2 A\cos^2 A} = \dfrac{1}{\sin^2 A\cos^2 A} \).
3. Using cos²A = 1 − sin²A, this is \( \dfrac{1}{\sin^2 A(1-\sin^2 A)} = \dfrac{1}{\sin^2 A - \sin^4 A} \).

Teacher's Note:
1. Write the answer in terms of sin A only to match the options.
2. Add the two fractions over the common denominator sin²A cos²A.

Answer: (b) \( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 2 \)

Explanation:
1. Square both equations: \( \dfrac{x^2}{a^2}\cos^2\theta + \dfrac{y^2}{b^2}\sin^2\theta + \dfrac{2xy}{ab}\sin\theta\cos\theta = 1 \) and \( \dfrac{x^2}{a^2}\sin^2\theta + \dfrac{y^2}{b^2}\cos^2\theta - \dfrac{2xy}{ab}\sin\theta\cos\theta = 1 \).
2. Add them; the cross terms cancel: \( \dfrac{x^2}{a^2}(\cos^2\theta+\sin^2\theta) + \dfrac{y^2}{b^2}(\sin^2\theta+\cos^2\theta) = 2 \).
3. Hence \( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 2 \).

Teacher's Note:
1. Squaring and adding removes θ when the two equations have the pattern cos/sin and sin/−cos.
2. The right side is 1 + 1 = 2, not 1.

Answer: (d) P: 1, Q: 1, R: 1

Explanation:
1. (i) (x/a)^2/3 = cos²θ and (y/b)^2/3 = sin²θ, so the sum is 1: P = 1.
2. (ii) The expression is sec²θ(cos²φ + sin²φ) − tan²θ = sec²θ − tan²θ = 1: Q = 1.
3. (iii) cos A = 1 − cos²A = sin²A, so sin²A + sin⁴A = cos A + cos²A = 1: R = 1.

Teacher's Note:
1. Each part reduces to sin² + cos² = 1 or sec² − tan² = 1.
2. In part (iii) replace sin²A by cos A.

Answer: (d) Both (a) and (b)

Explanation:
1. (a) cos(90° − θ) = sin θ, sec(90° − θ) = cosec θ, sin(90° − θ) = cos θ and cosec(90° − θ) = sec θ.
2. The expression becomes cos θ sin θ − sin³θ cos θ − cos³θ sin θ = cos θ sin θ − cos θ sin θ(sin²θ + cos²θ) = 0, so (a) is true.
3. (b) For A + B = 90°, cos A = sin B, so the root is \( \sqrt{1-\sin^2 B} = \cos B = \sin A \), so (b) is true.
4. Both statements are true: (d).

Teacher's Note:
1. Use the complementary-angle identities first and simplify afterwards.
2. When A + B = 90°, sin B = cos A and cos B = sin A.

Answer: (a) Only (i)

Explanation:
1. (i) Let s = sin θ cos θ. Then 3(1 − 2s)² + 6(1 + 2s) + 4(1 − 3s²) = 3 − 12s + 12s² + 6 + 12s + 4 − 12s² = 13, which is independent of θ.
2. (ii) a³ = cos²θ/sin θ and b³ = sin²θ/cos θ, so a²b² = (sin θ cos θ)^2/3 and a² + b² = (cos²θ + sin²θ)/(sin θ cos θ)^2/3 = 1/(sin θ cos θ)^2/3.
3. So a²b²(a² + b²) = 1, not 2: (ii) is false, and only (i) is correct.

Teacher's Note:
1. Use sin⁶θ + cos⁶θ = 1 − 3sin²θcos²θ and (sinθ − cosθ)² = 1 − 2sinθcosθ.
2. Test the statement (ii) with θ = 45°: a³ = b³ = 1/√2, which gives a²b²(a² + b²) = 1.

Answer: (a) 2

Explanation:
1. \( \sqrt{3+2\sqrt{2}} = \sqrt{(\sqrt{2}+1)^2} = \sqrt{2}+1 \), so the bracket is \( \sqrt{2} \) and \( (\sqrt{2})^4 = 4 \).
2. Hence cosec²θ = 1 + cot²θ = 4, so cosec θ = 2, sin θ = 1/2 and θ = 30° (acute).
3. The value is cot 30° + \( \dfrac{1/2}{1+\sqrt{3}/2} \) = √3 + \( \dfrac{1}{2+\sqrt{3}} \) = √3 + (2 − √3) = 2.

Teacher's Note:
1. Recognise 3 + 2√2 as the perfect square (√2 + 1)².
2. Another route: sin θ/(1 + cos θ) = (1 − cos θ)/sin θ = cosec θ − cot θ, so the value is cosec θ = 2.