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If $\sin ^{2} x+\sin ^{2} y=1$, then what is the value of $\cot (x+y)$ ?
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Given $\sin ^{2} x+\sin ^{2} y=1$
$\Rightarrow \sin ^{2} x=1-\sin ^{2} y$
$\Rightarrow \sin ^{2} x=\cos ^{2} y$
$\Rightarrow \sin x=\cos y$
Similarly By considering $\sin ^{2} y=1-\sin ^{2} x$, we have
$\cos x=\sin y$
Now, Consider $\cot (x+y)=\frac{\cos (x+y)}{\sin (x+y)}$
$=\frac{\cos x \cos y-\sin x \sin y}{\sin (x+y)}$
$=\frac{\cos x \cos y-\cos x \cos y}{\sin (x+y)}=0$
$\Rightarrow \sin ^{2} x=1-\sin ^{2} y$
$\Rightarrow \sin ^{2} x=\cos ^{2} y$
$\Rightarrow \sin x=\cos y$
Similarly By considering $\sin ^{2} y=1-\sin ^{2} x$, we have
$\cos x=\sin y$
Now, Consider $\cot (x+y)=\frac{\cos (x+y)}{\sin (x+y)}$
$=\frac{\cos x \cos y-\sin x \sin y}{\sin (x+y)}$
$=\frac{\cos x \cos y-\cos x \cos y}{\sin (x+y)}=0$
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