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Prove that:
$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right)$
$\sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right)$
$\sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
Solution:
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Verified Answer
L.H.S. $=\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)$
$-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$
Let $\frac{\pi}{4}-x=\mathrm{A}, \frac{\pi}{4}-y=\mathrm{B}$
L.H.S. $=\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}$
$=\cos (\mathrm{A}+\mathrm{B})$
$=\cos \left(\frac{\pi}{4}-x+\frac{\pi}{4}-y\right)$
$\begin{aligned}
&=\cos \left(\frac{\pi}{2}-(x+y)\right) \\
&=\sin (x+y)=\text { R.H.S. }
\end{aligned}$
$-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$
Let $\frac{\pi}{4}-x=\mathrm{A}, \frac{\pi}{4}-y=\mathrm{B}$
L.H.S. $=\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}$
$=\cos (\mathrm{A}+\mathrm{B})$
$=\cos \left(\frac{\pi}{4}-x+\frac{\pi}{4}-y\right)$
$\begin{aligned}
&=\cos \left(\frac{\pi}{2}-(x+y)\right) \\
&=\sin (x+y)=\text { R.H.S. }
\end{aligned}$
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