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  4. If sin-1 (2a/1+a2) + sin-1 (2b/1+b2) = 2 tan-1 x, then x is equal to :

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

Inverse Trigonometric Functions

Solution:

Let $a = \tan\theta b =\tan \phi$
$ \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[\sin2 \theta\right] $
$= 2 \theta = 2 \tan^{-1} a $
and $\sin^{-1} \left[\frac{2b}{1+b^{2}}\right] =\sin^{-1} \left[\frac{2\tan\phi}{1+ \tan^{2} \phi}\right] $
$= \sin^{-1} \left[\sin2 \phi\right] $
$ = 2\phi = 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} $