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Question: If angle \(\theta\) be divided into two parts such that the tangent of one part is \(k\) times the t...

If angle θ\theta be divided into two parts such that the tangent of one part is kk times the tangent of the other and φ\varphi is their difference, then sin⁡θ=\sin\theta =

A

k+1k−1sin⁡φ\frac{k + 1}{k - 1}\sin\varphi

B

k−1k+1sin⁡φ\frac{k - 1}{k + 1}\sin\varphi

C

2k−12k+1sin⁡φ\frac{2k - 1}{2k + 1}\sin\varphi

D

None of these

Answer

k+1k−1sin⁡φ\frac{k + 1}{k - 1}\sin\varphi

Explanation

Solution

Let A+B=θA + B = \theta andA−B=φA - B = \varphi.

Then tan⁡A=ktan⁡B\tan A = k\tan Bor k1=tan⁡Atan⁡B=sin⁡Acos⁡Bcos⁡Asin⁡B\frac{k}{1} = \frac{\tan A}{\tan B} = \frac{\sin A\cos B}{\cos A\sin B}

Applying componendo and dividendo

⇒k+1k−1=sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡Acos⁡B−cos⁡Asin⁡B\Rightarrow \frac{k + 1}{k - 1} = \frac{\sin A\cos B + \cos A\sin B}{\sin A\cos B - \cos A\sin B}

=sin⁡(A+B)sin⁡(A−B)=sin⁡θsin⁡φ⇒sin⁡θ=k+1k−1sin⁡φ= \frac{\sin(A + B)}{\sin(A - B)} = \frac{\sin\theta}{\sin\varphi} \Rightarrow \sin\theta = \frac{k + 1}{k - 1}\sin\varphi.