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$\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$ is equal to
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Verified Answer
The correct answer is:
$2 \sin x$
$\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$
$$
\begin{aligned}
&=\frac{2 \sin \left(\frac{x-3 x}{2}\right) \cos \left(\frac{x+3 x}{2}\right)}{-\left(\cos ^{2} x-\sin ^{2} x\right)} \\
&=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}=\frac{-2 \sin x \cos 2 x}{-\cos 2 x}=2 \sin x
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{2 \sin \left(\frac{x-3 x}{2}\right) \cos \left(\frac{x+3 x}{2}\right)}{-\left(\cos ^{2} x-\sin ^{2} x\right)} \\
&=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}=\frac{-2 \sin x \cos 2 x}{-\cos 2 x}=2 \sin x
\end{aligned}
$$
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