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Question: \[\frac{2\sin\theta\tan\theta(1 - \tan\theta) + 2\sin\theta\sec^{2}\theta}{(1 + \tan\theta)^{2}} =\]...

2sin⁡θtan⁡θ(1−tan⁡θ)+2sin⁡θsec⁡2θ(1+tan⁡θ)2=\frac{2\sin\theta\tan\theta(1 - \tan\theta) + 2\sin\theta\sec^{2}\theta}{(1 + \tan\theta)^{2}} =

A

sin⁡θ1+tan⁡θ\frac{\sin\theta}{1 + \tan\theta}

B

2sin⁡θ1+tan⁡θ\frac{2\sin\theta}{1 + \tan\theta}

C

2sin⁡θ(1+tan⁡θ)2\frac{2\sin\theta}{(1 + \tan\theta)^{2}}

D

None of these

Answer

2sin⁡θ1+tan⁡θ\frac{2\sin\theta}{1 + \tan\theta}

Explanation

Solution

Given expression =2sin⁡θ(1+tan⁡θ)2{tan⁡θ(1−tan⁡θ)+sec⁡2θ}= \frac{2\sin\theta}{(1 + \tan\theta)^{2}}\left\{ \tan\theta(1 - \tan\theta) + \sec^{2}\theta \right\}

=2sin⁡θ(1+tan⁡θ)2{tan⁡θ−tan⁡2θ+1+tan⁡2θ}=2sin⁡θ1+tan⁡θ= \frac{2\sin\theta}{(1 + \tan\theta)^{2}}\left\{ \tan\theta - \tan^{2}{}\theta + 1 + \tan^{2}\theta \right\} = \frac{2\sin\theta}{1 + \tan\theta}.