Question

Prove that $tan[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]+tan[\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}\frac{a}{b}]=\frac{2b}{a}$

Answer 1

The LHS part of the equation is:
$tan[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]+tan[\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}\frac{a}{b}]letco{s}^{-1}\frac{a}{b}=x\Rightarrow cosx=\frac{a}{b}tan[\frac{\pi }{4}+\frac{x}{2}]+tan[\frac{\pi }{4}-\frac{x}{2}]$
$=\frac{tan\frac{\pi }{4}+tan\frac{x}{2}}{1-tan\frac{\pi }{4}tan\frac{x}{2}}+\frac{tan\frac{\pi }{4}-tan\frac{x}{2}}{1+tan\frac{\pi }{4}.tan\frac{x}{2}}=\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}+\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}$
$=\frac{(1+tan\frac{x}{2}{)}^2+(1-tan\frac{x}{2}{)}^2}{(1-tan\frac{x}{2})(1+tan\frac{x}{2})}=\frac{2(1+ta{n}^2\frac{x}{2})}{1-ta{n}^2\frac{x}{2}}$
$=\frac{2}{cosx}$[since cos $2x=\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}]$
$=\frac{2b}{a}$