Question

$tan(\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b})+tan(\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}\frac{a}{b})$

Answer 1

$tan[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]+tan[\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}\frac{a}{b}]Letco{s}^{-1}\frac{a}{b}=x\Rightarrow cosx=\frac{a}{b}tan[\frac{\pi }{4}+\frac{x}{2}]+tan[\frac{\pi }{4}-\frac{x}{2}]=\frac{tan\frac{\pi }{4}+tan\frac{\pi }{2}}{1-tan\frac{\pi }{4}.tan\frac{\pi }{2}}+\frac{tan\frac{\pi }{4}-tan\frac{\pi }{2}}{1+tan\frac{\pi }{4}.tan\frac{\pi }{2}}=\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}+\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}=\frac{{(1-tan\frac{x}{2})}^2+{(1-tan\frac{x}{2})}^2}{(1-tan\frac{x}{2})(1+tan\frac{x}{2})}=\frac{2(1+ta{n}^2\frac{x}{2})}{1-ta{n}^2\frac{x}{2}}cos2x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}cosx=\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}2\times \frac{(1+ta{n}^2\frac{x}{2})}{1-ta{n}^2\frac{x}{2}}=\frac{2\times b}{a}$