Question
Show that $\cot ^{-1} \frac{1}{3}-\tan ^{-1} \frac{1}{3}=\cot ^{-1} \frac{3}{4}$.
$=\tan ^{-1} 3-\tan ^{-1} \frac{1}{3} \quad \cdots\left[\because \cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right)\right]$
$=\tan ^{-1}\left[\frac{3-\frac{1}{3}}{1+3\left(\frac{1}{3}\right)}\right]$
$=\tan ^{-1}\left[\frac{\left(\frac{8}{3}\right)}{1+1}\right]$
$=\tan ^{-1}\left(\frac{4}{3}\right)$
$\begin{aligned} & =\cot ^{-1}\left(\frac{3}{4}\right) \quad \cdots\left[\tan ^{-1} x=\cot ^{-1}\left(\frac{1}{x}\right)\right] \\ & =\text { RHS. }\end{aligned}$
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