Question

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

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

Answer 1

The correct option is D $-\sqrt{3}$

$2\sin (\theta +\frac{\pi }{3})=\cos (\theta -\frac{\pi }{6})$

$2[\sin \theta .\cos \frac{\pi }{3}+\cos \theta .\sin \frac{\pi }{3})=\cos \theta .\cos \frac{\pi }{6}+\sin \theta .\sin \frac{\pi }{6}$

$2[\frac{\sin \theta }{2}+\frac{\sqrt{3}\cos \theta }{2}]=\frac{\sqrt{3}}{2}\cos \theta +\frac{\sin \theta }{2}$

$2\sin \theta +2\sqrt{3}\cos \theta -\sqrt{3}\cos \theta -\sin \theta =0$

$\sin \theta +\sqrt{3}\cos \theta =0$

$\tan \theta =-\sqrt{3}$