Question

prove that : tan [π/4 + 1/2 cos^-1 a/b] + tan[π/4 - 1/2 cos^-1 a/b] = 2b/a

Answer 1

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Answer 2

Answer:

We have to Prove: $tan\left({{\pi\over4}+{1\over2}{{\cos}^{-1}}{a\over b}}\right)+tan\left({{\pi\over4}-{1\over2}{{\cos}^{-1}}{a \over b}}\right)={{2b} \over a}$

$={tan\left({{\pi\over 4}+{\theta\over2}}\right)}+{tan\left({{\pi\over4}-{\theta\over2}}\right)}$

$={{tan{\pi\over4}+tan{\theta\over2}}\over{1-tan{\pi\over4}\,tan{\theta\over2}}}+{{tan{\pi\over4}-tan{\theta \over2}}\over{1+tan{\pi\over4}\,tan{\theta\over2}}}$

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

$={{2\left( {1 + {{\tan }^2}{\theta \over 2}} \right)} \over {1 - {{\tan }^2}{\theta \over 2}}}\\\\\\= 2\left( {{1 \over {\cos \theta }}} \right)\\\\={{2b} \over a}\\\\= R.H.S.$

Hence Proved