Trigonometry & Inverse Trigonometry Question - None

Question

Question nos. 687 to 689

Column-1 represents a condition to form trigonometric equation. Column-2 represents the value of $\sin\theta + \cos\theta$ and Column-3 represents the general value of $\theta$ satisfying the trigonometric equation.

Column-1	Column-2	Column-3
(I) If $2^{\sin\theta}$, $\sqrt{2}$ and $2^{\cos\theta}$ are three terms of a decreasing G.P.	(i) $\dfrac{\sqrt{3}+1}{2}$	(P) $\theta = 2n\pi - \dfrac{\pi}{2}$
(II) If $\cos\theta$, $\sec\theta$ and $\cot\theta$ are three positive numbers in H.P.	(ii) $\sqrt{2}$	(Q) $\theta = 2n\pi + \dfrac{\pi}{6}$
(III) If $2\log\sec\theta$, $\log 2$ and $2\log\text{cosec}\,\theta$ are in A.P.	(iii) $-1$	(R) $\theta = 2n\pi + \dfrac{\pi}{2}$
(IV) If G.M. of $(2+\sin\theta)$, $(3+\sin\theta)$ and $(4+\sin\theta)$ is equal to cube root of 6.	(iv) $1$	(S) $\theta = 2n\pi + \dfrac{\pi}{4}$

689. Which of the following options is the only correct combination?

Accepted Answer

Step 1: Condition (IV) - Geometric Mean of (2+sin θ), (3+sin θ), (4+sin θ)

The GM equals the cube root of 6:

∛[(2+sin θ)(3+sin θ)(4+sin θ)] = ∛6

(2+sin θ)(3+sin θ)(4+sin θ) = 6

Let sin θ = x. Then: (2+x)(3+x)(4+x) = 6

Expanding: x³ + 9x² + 26x + 24 = 6

x³ + 9x² + 26x + 18 = 0

Testing x = -1: (-1)³ + 9(-1)² + 26(-1) + 18 = -1 + 9 - 26 + 18 = 0 ✓

So sin θ = -1

Step 2: Find cos θ when sin θ = -1

Since sin²θ + cos²θ = 1 and sin θ = -1:

1 + cos²θ = 1 → cos θ = 0

Step 3: Calculate sin θ + cos θ

sin θ + cos θ = -1 + 0 = -1 (iii)

Step 4: Find the general solution

sin θ = -1 occurs when θ = 2nπ - π/2 (P)

Step 5: Verify the match

Condition (IV) matches with:

Column-2: (iii) -1

Column-3: (P) θ = 2nπ - π/2

Therefore: (IV) (iii) (P)

∴ Answer: D

Question:

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