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The value of $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)$
$-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$ is equal to
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$-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$ is equal to
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Verified Answer
The correct answer is:
$\sin (x+y) \quad$
We have $\cos A \cos B-\sin A \sin B$
$=\cos (A+B)=\cos \left[\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right]$
where, $A=\frac{\pi}{4}-x$ and $B=\frac{\pi}{4}-y$
$=\cos \left[\frac{\pi}{2}-(x+y)\right]=\sin (x+y)$
$=\cos (A+B)=\cos \left[\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right]$
where, $A=\frac{\pi}{4}-x$ and $B=\frac{\pi}{4}-y$
$=\cos \left[\frac{\pi}{2}-(x+y)\right]=\sin (x+y)$
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