Question
Prove the following:
$\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{4}$

Answer

L.H.S. $=\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)$
$=\tan ^{-1}\left[\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right]$
$=\tan ^{-1}\left(\frac{3+2}{6-1}\right)$
$=\tan ^{-1}(1)$
$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$
$=\frac{\pi}{4}$
= R.H.S.

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