Question

Prove that $2{\tan }^{-1}\left[\sqrt{\frac{a-b}{a+b}}tan\frac{x}{2}\right]={\cos }^{-1}\left[\frac{b+a\cos x}{a+b\cos x}\right]$

Answer 1

Prove that $2{\tan }^{-1}\left[\frac{\sqrt{a-b}}{\sqrt{a+b}}\tan \frac{x}{2}\right]={\cos }^{-1}\left[\frac{a\cos a+b}{a+b\cos a}\right]$

$LHS=2{\tan }^{-1}[\frac{\sqrt{a-b}}{\sqrt{a+b}}.\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}]$

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

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

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

$={\cos }^{-1}\left[\frac{a\cos x+b}{a+b\cos x}\right]$ $(\because {\cos }^2\theta -{\sin }^2\theta \Rightarrow \cos x={\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})$