Question

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

Answer 1

$\begin{array}{c}=2\left[\sin \theta \cos \frac{\pi }{3}+\cos \theta \sin \frac{\pi }{3}\right]=\left(\cos \theta \cdot \cos \frac{\pi }{6}+\sin \theta \sin \frac{\pi }{6}\right)\\ =2\left[\sin \theta \left(\frac{1}{2}\right)+\cos \theta \left(\frac{\sqrt{3}}{2}\right)\right]=\left[\frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta \right]\\ =\sin \theta +\sqrt{3}\cos \theta =\frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta \\ =\sqrt{3}\cos \theta -\frac{\sqrt{3}}{2}\cos \theta =\frac{1}{2}\sin \theta -\sin \theta \\ =\sqrt{3}\left(\frac{2\cos -\cos \theta }{2}\right)=\frac{\sin \theta -2\sin \theta }{2}\\ =\sqrt{3}\cos \theta =-\sin \theta \\ =\frac{-\sin \theta }{\cos \theta }=\sqrt{3}\\ =-\tan \theta =\sqrt{3}\Rightarrow \sqrt{3}+\tan \theta =0\end{array}$

Hence, proved.