Question

Prove that
If $2\sin (\theta +\frac{\pi }{3})=\cos (\theta -\frac{\pi }{6})then,\tan \theta +\sqrt{3}=0$

Answer 1

$\sin (A+B)=\sin A\cos B+\sin B\cos A$

$\cos (A-B)=\cos A\cos B-\sin A\sin B$

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

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

$\to 2(\frac{\sin \theta }{2}+\frac{\sqrt{3}\cos \theta }{2})=\frac{\sqrt{3}\cos \theta }{2}+\frac{\sin \theta }{2}$

$\to 2(\sin \theta +\sqrt{3}\cos \theta )=\sqrt{3}\cos \theta +\sin \theta$

$\to \sin \theta =-\sqrt{3}\cos \theta$

$\to \frac{\sin \theta }{\cos \theta }=-\sqrt{3}$

$\to \tan \theta +\sqrt{3}=0$