If 2 sin left(theta+frac{pi}{3}right)=cos lef | vedclass

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(B) Given: $2 \sin \left(	heta+\frac{\pi}{3}ight)=\cos \left(	heta-\frac{\pi}{6}ight)$ Using the expansion formulas $\sin(A+B) = \sin A \cos B +

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$-\sqrt{3}$

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(B) Given: $2 \sin \left(	heta+\frac{\pi}{3}ight)=\cos \left(	heta-\frac{\pi}{6}ight)$ Using the expansion formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$: $2 \left[\sin 	heta \cos \frac{\pi}{3} + \cos 	heta \sin \frac{\pi}{3}ight] = \cos 	heta \cos \frac{\pi}{6} + \sin 	heta \sin \frac{\pi}{6}$ Substituting the values $\cos \frac{\pi}{3} = \frac{1}{2}$,$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,and $\sin \frac{\pi}{6} = \frac{1}{2}$: $2 \left(\frac{1}{2} \sin 	heta + \frac{\sqrt{3}}{2} \cos 	hetaight) = \frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta$ $\sin 	heta + \sqrt{3} \cos 	heta = \frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta$ Subtracting $\frac{1}{2} \sin 	heta$ and $\frac{\sqrt{3}}{2} \cos 	heta$ from both sides: $\frac{1}{2} \sin 	heta + \frac{\sqrt{3}}{2} \cos 	heta = 0$ $\sin 	heta = -\sqrt{3} \cos 	heta$ Therefore,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = -\sqrt{3}$.

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

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