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