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

L.H.S. {tex}\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]{/tex}

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

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

= {tex}{{\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}}}{/tex}

= {tex}{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}} + {{1 - tan{\theta \over 2}} \over {1 + \tan {\theta \over 2}}}{/tex}

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

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

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

= {tex}{{2b} \over a}{/tex}

= R.H.S.