Let $\theta=\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)$
then L.H.S. $=\tan \left[\frac{\pi}{4}+\theta\right]+\tan \left[\frac{\pi}{4}-\theta\right]$
$=\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \cdot \tan \theta}+\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \cdot \tan \theta}$
$=\frac{1+\tan \theta}{1-\tan \theta}+\frac{1-\tan \theta}{1+\tan \theta}=\frac{(1+\tan \theta)^2+(1-\tan \theta)^2}{(1-\tan \theta)(1+\tan \theta)}$
$=\frac{1+2 \tan \theta+\tan ^2 \theta+1-2 \tan \theta+\tan ^2 \theta}{\left(1-\tan ^2 \theta\right)}$
$=\frac{2+2 \tan ^2 \theta}{\left(1-\tan ^2 \theta\right)}=\frac{2\left(1+\tan ^2 \theta\right)}{\left(1-\tan ^2 \theta\right)}=\frac{2}{\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}}=\frac{2}{\cos 2 \theta}$
$=\frac{2}{\cos \left[2 \times \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]} \quad\left[\because \theta=\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$
$=\frac{2}{\cos \left(\cos ^{-1} \frac{a}{b}\right)}=\frac{2}{\frac{a}{b}}=\frac{2 b}{a}=\text { R.H.S. }$
