Question

Prove that $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$

Answer 1

L.H.S. $=\tan ^{-1} x+\cot ^{-1} x$

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

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

$=\tan ^{-1}\left(\frac{x^2+1}{0}\right)=\tan ^{-1}(\infty)=\frac{\pi}{2}=\text { R.H.S. }$