Question

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

• A. ${\cos }^{-1}\left(\frac{b+a\cos x}{a+b\cos x}\right)$
• B. ${\cos }^{-1}\left(\frac{b+a\cos x}{a-b\cos x}\right)$
• C. ${\cos }^{-1}\left(\frac{b-a\cos x}{a+b\cos x}\right)$
• D. ${\cos }^{-1}\left(\frac{b-a\cos x}{a-b\cos x}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{b+a\cos x}{a+b\cos x}\right)$

Given

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

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

$⟹\tan \theta =\left(\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2}\right)$

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

$⟹{\tan }^2\theta =\frac{a-b}{a+b}{\tan }^2\frac{x}{2}\cdots \left(2\right)$

$\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\cdots \left(3\right)$

From $\left(2\right)$

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

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

$⟹\cos 2\theta =\frac{a+b-a{\tan }^2\frac{x}{2}+b{\tan }^2\frac{x}{2}}{a+b+a{\tan }^2\frac{x}{2}-b{\tan }^2\frac{x}{2}}$

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

Dividing Numerator and Denominator by $(1+{\tan }^2\frac{x}{2})$

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

$⟹\cos 2\theta =\frac{a\cos x+b}{a+b\cos x}$

$(\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta })$

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

From $\left(1\right)$

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