If theta is any angle,then sin^2 theta cos^2 | vedclass

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(D) We know that $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$.Substituting these into the expression:

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$\frac{1}{8}(1 - \cos 4	heta)$

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(D) We know that $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$.Substituting these into the expression:$\sin^2 	heta \cos^2 	heta = \left(\frac{1 - \cos 2	heta}{2}ight) \left(\frac{1 + \cos 2	heta}{2}ight)$$= \frac{(1 - \cos 2	heta)(1 + \cos 2	heta)}{4}$$= \frac{1 - \cos^2 2	heta}{4}$Using the identity $\cos^2 A = \frac{1 + \cos 2A}{2}$,we have $\cos^2 2	heta = \frac{1 + \cos 4	heta}{2}$.$= \frac{1 - \left(\frac{1 + \cos 4	heta}{2}ight)}{4}$$= \frac{\frac{2 - 1 - \cos 4	heta}{2}}{4}$$= \frac{1 - \cos 4	heta}{8}$

If $\theta$ is any angle,then $\sin^2 \theta \cos^2 \theta =$

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