Solveeit Logo
Login

Question

Question: \[\frac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} =\]...

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

A

12tan⁡θ\frac{1}{2}\tan\theta

B

12cot⁡θ\frac{1}{2}\cot\theta

C

tan⁡θ\tan\theta

D

cot⁡θ\cot\theta

Answer

tan⁡θ\tan\theta

Explanation

Solution

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

=sin⁡θ+2sin⁡θcos⁡θ2cos⁡2θ+cos⁡θ=sin⁡θ(1+2cos⁡θ)cos⁡θ(1+2cos⁡θ)=tan⁡θ= \frac{\sin\theta + 2\sin\theta\cos\theta}{2\cos^{2}\theta + \cos\theta} = \frac{\sin\theta(1 + 2\cos\theta)}{\cos\theta(1 + 2\cos\theta)} = \tan\theta.

Trick : Put θ=30∘\theta = 30{^\circ}, since for θ=30∘\theta = 30{^\circ}no option will give the common value.