Tan(π/4+1/2cos-¹a/b)+tan(π/4-1/2cos-¹a/b)=2b/a Prove that

L.H.S. $\tan \left[ {{\pi \over 4} + {1 \over 2}{{\cos }^{ - 1}}{a \over b}} \right] + \tan \left[ {{\pi \over 4} - {1 \over 2}{{\cos }^{ - 1}}{a \over b}} \right]$

Let ${\cos ^{ - 1}}{a \over b} = \theta ,$ then $\cos \theta = {a \over b}$ and ${1 \over 2}{\cos ^{ - 1}}{a \over b} = {\theta \over 2}$

= $\tan \left[ {{\pi \over 4} + {\theta \over 2}} \right] + \tan \left[ {{\pi \over 4} - {\theta \over 2}} \right]$

= ${{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4}\tan {\theta \over 2}}} + {{\tan {\pi \over 4} - tan{\theta \over 2}} \over {1 + \tan {\pi \over 4}\tan {\theta \over 2}}}$

= ${{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}} + {{1 - tan{\theta \over 2}} \over {1 + \tan {\theta \over 2}}}$

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

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

= $2\left( {{1 \over {\cos \theta }}} \right)$

= ${{2b} \over a}$

= R.H.S.