Prove that:

To Prove: $\tan ^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{x y}}\right)=\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{y}$

We know that, $\tan A+\tan B=\frac{\tan A+\tan B}{1-\tan A \tan B}$

Also, $\tan ^{-1}\left(\frac{\mathrm{A}+\mathrm{B}}{1-\mathrm{AB}}\right)=\tan ^{-1} \mathrm{~A}+\tan ^{-1} \mathrm{~B}$

Taking $A=\sqrt{x}$ and $B=\sqrt{y}$

We get,

$\tan ^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{x y}}\right)=\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{y}$

Hence, Proved.