frac{1}{4} [sqrt{3} cos 23^circ - sin 23^circ | vedclass

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Question

(D) Given expression: $\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ]$Multiply and divide by $2$ inside the bracket:$= \frac{1}{2} [\frac{\sqrt{

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given expression: $\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ]$Multiply and divide by $2$ inside the bracket:$= \frac{1}{2} [\frac{\sqrt{3}}{2} \cos 23^\circ - \frac{1}{2} \sin 23^\circ]$Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$:$= \frac{1}{2} [\cos 30^\circ \cos 23^\circ - \sin 30^\circ \sin 23^\circ]$Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:$= \frac{1}{2} \cos(30^\circ + 23^\circ)$$= \frac{1}{2} \cos 53^\circ$Since $\frac{1}{2} \cos 53^\circ$ is not equal to any of the given options,the correct answer is $D$.

$\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ] = $

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