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  4. ( sin x- sin 3 x/ sin 2 x- cos 2 x) is equal to

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Q. $\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}$ is equal to

Trigonometric Functions

Solution:

$ \frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x} =\frac{\sin 3 x-\sin x}{\cos ^2 x-\sin ^2 x} $
$ =\frac{2 \cos \frac{3 x+x}{2} \cdot \sin \frac{3 x-x}{2}}{\cos 2 x} $
$ =\frac{2 \cos 2 x \cdot \sin x}{\cos 2 x} $
$ =2 \sin x$