Trigonometry: JEE Advanced PYQ, 2007–2026

Answer: (D)

tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹((1/2 + 1/3)/(1 − 1/6)) = tan⁻¹((5/6)/(5/6)) = tan⁻¹ 1 = π/4.

Answer: (A)

Squaring and adding: (sin²α + cos²α) + (sin²β + cos²β) + 2(sin α sin β + cos α cos β) = 1/4 + 9/4. So 2 + 2 cos(α − β) = 5/2, and cos(α − β) = 1/4.

Answer: (A)

g(x) = sin⁻¹(sin x) is a zigzag of period 2π with values in [−π/2, π/2]. It equals x on [−π/2, π/2], π − x on [π/2, 3π/2], x − 2π on [3π/2, 5π/2], and so on. A solution needs |x/10| ≤ π/2, so |x| ≤ 5π ≈ 15.7. For x > 0, on [0, π/2] we have x = x/10 only at x = 0. The arms that follow give: π − x = x/10 → x ≈ 2.86, in [π/2, 3π/2]; x − 2π = x/10 → x ≈ 6.98, in [3π/2, 5π/2]; 3π − x = x/10 → x ≈ 8.57, in [5π/2, 7π/2]; x − 4π = x/10 → x ≈ 13.96, in [7π/2, 9π/2]; 5π − x = x/10 → x ≈ 14.28, in [9π/2, 11π/2]. That is 5 positive roots. The next arm is rising, but the line is already above π/2 there, so there are no more roots. By oddness there are 5 negative roots, plus x = 0, giving 11.

Answer: (A)

Rewrite the condition as sinθ = 1 − sin²θ = cos²θ. Squaring, sin²θ = cos⁴θ. Therefore cos²θ + cos⁴θ = sinθ + sin²θ, which is 1 by the given condition. No need to solve the quadratic.

Answer: (A)

Let S = cos(π/7) + cos(3π/7) + cos(5π/7). Multiply by 2 sin(π/7) and use 2 sin A cos B = sin(A+B) − sin(B−A) term by term. 2 sin(π/7)cos(π/7) = sin(2π/7). 2 sin(π/7)cos(3π/7) = sin(4π/7) − sin(2π/7). 2 sin(π/7)cos(5π/7) = sin(6π/7) − sin(4π/7). Adding, everything cancels except sin(6π/7), so 2 sin(π/7)·S = sin(6π/7) = sin(π − π/7) = sin(π/7). Since sin(π/7) ≠ 0, S = 1/2.

Answer: (D)

Write the expression as (cos 10° − √3 sin 10°)/(sin 10° cos 10°). In the numerator, factor out 2: 2[cos 10° cos 60° − sin 10° sin 60°] = 2 cos 70° = 2 sin 20°. The denominator is ½ sin 20°. The ratio is 2 sin 20° / (½ sin 20°) = 4.

Answer: (A)

Principle: tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)) only when xy < 1. If xy > 1, the sum differs by ±π. Apply: take the tangent of both sides, giving 3x/(1 − 2x²) = 1, so 2x² + 3x − 1 = 0 and x = (−3 ± √17)/4. For x = (−3 + √17)/4 ≈ 0.28, 2x² < 1 and the sum is tan⁻¹(1) = π/4, which is valid. For x = (−3 − √17)/4 ≈ −1.78, both inverse tangents are negative and 2x² > 1, so the sum is π/4 − π = −3π/4. This root is rejected. The only solution, and hence the required sum, is (√17 − 3)/4.

Answer: (B)

Multiplying by 8 sin 20°: 8 sin 20° cos 20° cos 40° cos 80° = 4 sin 40° cos 40° cos 80° = 2 sin 80° cos 80° = sin 160° = sin 20°. So the product is 1/8.

Answer: (D)

From sin x + sin²x = 1 we get sin x = 1 − sin²x = cos²x. Squaring, sin²x = cos⁴x, so cos²x + cos⁴x = sin x + sin²x = 1. Now factor the expression: cos¹²x + 3cos¹⁰x + 3cos⁸x + cos⁶x = cos⁶x(cos⁶x + 3cos⁴x + 3cos²x + 1) = cos⁶x(1 + cos²x)³ = (cos²x + cos⁴x)³ = 1. Hence the value is 1 − 2 = −1.

Answer: (C)

In the third quadrant sin θ = −√(1 − 9/25) = −4/5. Then tan(θ/2) = sin θ/(1 + cos θ) = (−4/5)/(2/5) = −2. The sign agrees with θ/2 lying in the second quadrant.