Question

If $\tan \left(\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right)+\tan \left(\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right)=\frac{mb}{a}$.Find $m$

Answer 1

L.H.S

$\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\pm \tan A\tan B}$

$\frac{1+\tan \left(\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right)}{1-\tan \left(\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)\right)}+\frac{1-\tan \left(\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)\right)}{1+\tan \left(\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)\right)}$

$\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x},x=\frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)$

$=\frac{{\left(1+\tan \right)}^2+{\left(1-\tan x\right)}^2}{1-{\tan }^{2}x}$

$=2\frac{\left(1+{\tan }^{2}x\right)}{1-{\tan }^{2x}}=\frac{2{\sec }^{2}x}{1-{\tan }^{2}x}$

$=\frac{2}{{\cos }^{2}x\left(1-\frac{{\sin }^{2}x}{co{x}^{2}x}\right)}=\frac{2}{{\cos }^{2}x-{\sin }^{2}x}=\frac{2}{\cos 2x}=\frac{2b}{a}$

$\therefore m=2$