Prove that $\tan \left[\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)\right]=\frac{2b}{a}$

Take 

$\tan \left[\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)\right]$

Put $\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)=\theta \Rightarrow {\cos }^{-1}\left(\frac{a}{b}\right)=2\theta$

$\Rightarrow \cos 2\theta =\frac{a}{b}$

$=\tan \left[\frac{\pi }{4}+\theta \right]+\tan \left[\frac{\pi }{4}-\theta \right]=\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }$

$=\frac{(1+\tan \theta {)}^2+(1-\tan \theta {)}^2}{(1-\tan \theta )(1+\tan \theta )\mid }$

$=\frac{1+{\tan }^2\theta +2\tan \theta +1+{\tan }^2\theta -2\tan \theta }{1-{\tan }^2\theta }$

$=\frac{2+2{\tan }^2\theta }{1-{\tan }^2\theta }=2\left(\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }\right)$

$=\frac{2}{\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }}=\frac{2}{\cos 2\theta }=\frac{2}{\left(\frac{a}{b}\right)}=\frac{2b}{a}$