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If $\theta$ is any angle, then $\sin ^2 \theta \cos ^2 \theta=$
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The correct answer is:
$\frac{1}{8}(1-\cos 4 \theta)$
$\begin{aligned} & \text { Here, } \sin ^2 \theta \cos ^2 \theta \\ & =\left(\frac{1-\cos 2 \theta}{2}\right)\left(\frac{1+\cos 2 \theta}{2}\right) \\ & =\frac{1-\cos ^2 2 \theta}{4}=\frac{1-\left(\frac{1+\cos 4 \theta}{2}\right)}{4} \\ & =\frac{1-\frac{1}{2}-\frac{\cos 4 \theta}{2}}{4}=\frac{1}{8}(1-\cos 4 \theta)\end{aligned}$
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