frac{sqrt{3} sin theta + cos theta}{sin left( | vedclass

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Question

(C) We are given the expression: $\frac{\sqrt{3} \sin 	heta + \cos 	heta}{\sin \left(	heta + \frac{\pi}{6}ight)}$.Using the expansion formula $\s

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We are given the expression: $\frac{\sqrt{3} \sin 	heta + \cos 	heta}{\sin \left(	heta + \frac{\pi}{6}ight)}$.Using the expansion formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we expand the denominator:$\sin \left(	heta + \frac{\pi}{6}ight) = \sin 	heta \cos \frac{\pi}{6} + \cos 	heta \sin \frac{\pi}{6}$.Since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{\pi}{6} = \frac{1}{2}$,the denominator becomes:$\frac{\sqrt{3}}{2} \sin 	heta + \frac{1}{2} \cos 	heta = \frac{1}{2} (\sqrt{3} \sin 	heta + \cos 	heta)$.Substituting this back into the original expression:$\frac{\sqrt{3} \sin 	heta + \cos 	heta}{\frac{1}{2} (\sqrt{3} \sin 	heta + \cos 	heta)} = \frac{1}{1/2} = 2$.Thus,the correct option is $C$.

$\frac{\sqrt{3} \sin \theta + \cos \theta}{\sin \left(\theta + \frac{\pi}{6}\right)} = $

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