Question

State whether True/False

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

• A. True
• B. False

Answer 1

The correct option is B False

We know that

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

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

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

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

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

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

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

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

We know that

$\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\cos x$

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

Hence False.