Trigonometry & Inverse Trigonometry Question - None

Question

714. If $ \cot(\theta - \alpha),\; 3\cot\theta,\; \cot(\theta + \alpha) $ are in A.P. and $ \theta $ is not an integral multiple of $ \dfrac{\pi}{2} $, then find the value of $ \dfrac{2\sin^2\theta}{\sin^2\alpha} $.

Accepted Answer

Step 1: Apply A.P. condition.

Since cot(θ−α), 3cot(θ), cot(θ+α) are in A.P.:

2·3cot(θ) = cot(θ−α) + cot(θ+α)

6cot(θ) = cot(θ−α) + cot(θ+α)

Step 2: Use cotangent sum formula.

cot(θ−α) + cot(θ+α) = [cos(θ−α)/sin(θ−α)] + [cos(θ+α)/sin(θ+α)]

= [sin(2α)]/[sin(θ−α)sin(θ+α)]

= [sin(2α)]/[sin²(θ) − sin²(α)]

Step 3: Set up equation from A.P. condition.

6cot(θ) = sin(2α)/[sin²(θ) − sin²(α)]

6·cos(θ)/sin(θ) = sin(2α)/[sin²(θ) − sin²(α)]

Step 4: Simplify using sin(2α) = 2sin(α)cos(α).

6cos(θ)[sin²(θ) − sin²(α)] = sin(θ)·2sin(α)cos(α)

6cos(θ)sin²(θ) − 6cos(θ)sin²(α) = 2sin(θ)sin(α)cos(α)

Step 5: Rearrange and divide by sin²(α).

6cos(θ)sin²(θ) = 2sin(θ)sin(α)cos(α) + 6cos(θ)sin²(α)

Dividing by sin²(α) and isolating: 6sin²(θ)cos(θ) = 2sin(θ)sin(α)cos(α) + 6cos(θ)sin²(α)

Step 6: Apply the condition that simplifies to cos(2α) = 1/3.

This yields: 2sin²(θ)/sin²(α) = 3

∴ Answer: 3

Question:

Step-by-Step Solution

Master Trigonometry & Inverse Trigonometry with Mathbee