Question

Prove:

$2{tan}^{-1}$ $\left(\sqrt{\frac{a-b}{a+b}}tan\frac{\theta }{2}\right)={cos}^{-1}$ $\left(\frac{acos\theta +b}{a+bcos\theta }\right)$

Answer 1

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

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

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

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

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

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

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

$={\cos }^{-1}\left\{\frac{a\cos x+b}{a+b\cos x}\right\}$

$=$R.H.S

Hence proved.