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

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

Vedclass · Accepted Answer

$-\sqrt{3}$

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(D) Given the equation: $2 \sin \left(	heta+\frac{\pi}{3}ight)=\cos \left(	heta-\frac{\pi}{6}ight)$ Using the expansion formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$: $2 \left(\sin 	heta \cos \frac{\pi}{3} + \cos 	heta \sin \frac{\pi}{3}ight) = \cos 	heta \cos \frac{\pi}{6} + \sin 	heta \sin \frac{\pi}{6}$ Substitute 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 $\sin \frac{\pi}{6} = \frac{1}{2}$: $2 \left(\sin 	heta \cdot \frac{1}{2} + \cos 	heta \cdot \frac{\sqrt{3}}{2}ight) = \cos 	heta \cdot \frac{\sqrt{3}}{2} + \sin 	heta \cdot \frac{1}{2}$ $\sin 	heta + \sqrt{3} \cos 	heta = \frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta$ Multiply by $2$ to clear the fractions: $2 \sin 	heta + 2 \sqrt{3} \cos 	heta = \sqrt{3} \cos 	heta + \sin 	heta$ Rearrange the terms: $2 \sin 	heta - \sin 	heta = \sqrt{3} \cos 	heta - 2 \sqrt{3} \cos 	heta$ $\sin 	heta = -\sqrt{3} \cos 	heta$ Divide by $\cos 	heta$: $	an 	heta = -\sqrt{3}$

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

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