Question:
Write the value of $\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right)$ for $x>0$
Solution:
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right), x y<1$
$\therefore \tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\tan ^{-1}\left(\frac{x+\frac{1}{x}}{1-x \frac{1}{x}}\right), x>0=$
$=\tan ^{-1}\left(\frac{x^{2}+1}{0}\right)$
$=\tan ^{-1}(\infty)$
$=\tan ^{-1}\left(\tan \frac{\pi}{2}\right)$
$=\frac{\pi}{2}$
$\therefore \tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\frac{\pi}{2}$
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