Question:
Prove that
$\tan ^{-1}\left(\frac{x+\sqrt{x}}{1-x^{3 / 2}}\right)=\tan ^{-1} x+\tan ^{-1} \sqrt{x}$
Solution:
We know that,
$\tan ^{-1}\left(\frac{A+B}{1-A B}\right)=\tan ^{-1} A+\tan ^{-1} B$
Now, taking $A=x$ and $B=\sqrt{x}$
We get,
$\tan ^{-1} x+\tan ^{-1} \sqrt{x}=\tan ^{-1}\left(\frac{x+\sqrt{x}}{1-x^{3 / 2}}\right)$
As, $x \cdot x^{1 / 2}=x^{3 / 2}$
Hence, Proved.