Question

If ${\sin }^{-1}\frac{2a}{1+a^2}+{\sin }^{-1}\frac{2b}{1+b^2}=2{\tan }^{-1}x,$ then x is equal to :

• A. $\frac{a-b}{1+ab}$
• B. $\frac{b}{1+ab}$
• C. $\frac{b}{1-ab}$
• D. $\frac{a+b}{1-ab}$

Answer 1

Answer: (D)

Let $a=\tan \theta b=\tan \varphi$
$\therefore {\sin }^{-1}\left[\frac{2a}{1+a^2}\right]={\sin }^{-1}\left[\frac{2\tan \theta }{1+{\tan }^2\theta }\right]$
$={\sin }^{-1}\left[\sin 2\theta \right]$
$=2\theta =2{\tan }^{-1}a$
and ${\sin }^{-1}\left[\frac{2b}{1+b^2}\right]={\sin }^{-1}\left[\frac{2\tan \varphi }{1+{\tan }^2\varphi }\right]$
$={\sin }^{-1}\left[\sin 2\varphi \right]$
$=2\varphi =2{\tan }^{-1}b$
Thus, ${\sin }^{-1}\left[\frac{2a}{1+a^2}\right]=2{\tan }^{-1}a$ and
${\sin }^{-1}\left[\frac{2b}{1+b^2}\right]=2{\tan }^{-1}b$
$\therefore 2{\tan }^{-1}x={\sin }^{-1}\left[\frac{2a}{1+a^2}\right]+{\sin }^{-1}\left[\frac{2b}{1+b^2}\right]$
$=2{\tan }^{-1}a+2{\tan }^{-1}b$
$\Rightarrow {\tan }^{-1}x={\tan }^{-1}a+{\tan }^{-1}b$
${\tan }^{-1}x={\tan }^{-1}\frac{a+b}{1-ab}$
$\therefore x=\frac{a+b}{1-ab}$