If $2 \sin \left( heta+\frac{\pi}{3} ight)=\cos \left( heta-\frac{\pi}{6} ight)$, then $ an heta$

Question

If $2 \sin \left(	heta+\frac{\pi}{3}ight)=\cos \left(	heta-\frac{\pi}{6}ight)$, then $	an 	heta$

Aicharya · Accepted Answer

$\begin{aligned} & 2 \sin \left(	heta+\frac{\pi}{3}ight)=\cos \left(	heta-\frac{\pi}{6}ight) \ & 2\left[\sin 	heta \cos \frac{\pi}{3}+\cos 	heta \sin \frac{\pi}{3}ight]=\left(\cos 	heta \cos \frac{\pi}{6}+\sin 	heta \sin \frac{\pi}{6}ight) \ & 	herefore 2\left(\frac{\sin 	heta}{2}+\frac{\sqrt{3} \cos 	heta}{2}ight)=\left(\frac{\sqrt{3} \cos 	heta}{2}+\frac{\sin 	heta}{2}ight) \ & 	herefore\left(\frac{\sin 	heta}{2}+\frac{\sqrt{3} \cos 	heta}{2}ight)=0 \Rightarrow \sin 	heta+\sqrt{3} \cos 	heta=0 \Rightarrow 	an 	heta=-\sqrt{3} \end{aligned}$

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

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