Question

If $2 \sin \left( \theta + \frac{\pi}{3}\right) = \cos \left( \theta -\frac{\pi}{6}\right) , $ then $\tan \, \theta = $

• A. $\sqrt{3}$
• B. $- \frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $- \sqrt{3}$

Answer 1

Answer: (D)

We have, $ 2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right) $
$\Rightarrow 2\left[\sin \theta \cos \frac{\pi}{3}+\cos \theta \sin \frac{\pi}{3}\right] $
$= \cos \theta \cos \frac{\pi}{6}+\sin \theta \sin \frac{\pi}{6} $
$ \Rightarrow 2\left[\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right]=\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta $
$ \Rightarrow 2 \sin \theta+2 \sqrt{3} \cos \theta=\sqrt{3} \cos \theta+\sin \theta $
$\Rightarrow \sin \theta=-\sqrt{3} \cos \theta $
$\Rightarrow \tan \theta=-\sqrt{3} $