Question:
If $\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}$, then write the value of $x+y+x y .$
Solution:
We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Now,
$\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}$
$\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4}$
$\Rightarrow \frac{x+y}{1-x y}=\tan \frac{\pi}{4}$
$\Rightarrow \frac{x+y}{1-x y}=1$
$\Rightarrow x+y=1-x y$
$\Rightarrow x+y+x y=1$
$\therefore x+y+x y=1$
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