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Prove $: \tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)=\cos ^{-1}\left(\frac{9}{2}\right)$
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Solution:

$\begin{aligned} L \cdot H \cdot S \cdot &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right) \\ &=\tan ^{-1}\left[\frac{\frac{1}{7}+\frac{1}{13}}{1-\left(\frac{1}{7}\right)\left(\frac{1}{13}\right)}\right] \\ & =\tan ^{-1}\left(\frac{2}{9}\right) \\ & R.H.S.=\cot ^{-1}\left(\frac{9}{2}\right)\\ & \cot ^{-1}\left(\frac{9}{2}\right) \neq \cos ^{-1}\left(\frac{9}{2}\right)\\ & \therefore L . H . S . \neq R . H . S \end{aligned}$

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