STD 12 - 2. inverse trigonometric function — JEE कक्षा 11 साइन्स — Question

${\sin ^{ - 1}}\frac{{\sqrt x }}{{\sqrt {x + a} }} = {\sin ^{ - 1}}\frac{{\sqrt a \sqrt {{{\tan }^2}\theta } }}{{\sqrt {a\,{{\tan }^2}\theta + a} }} $

$= {\sin ^{ - 1}}\frac{{\sqrt a \,\tan \theta }}{{\sqrt a \,\sec \theta }}$

$ = {\sin ^{ - 1}}\sin \theta = \theta = {\tan ^{ - 1}}\left( {\sqrt {\frac{x}{a}} } \right)$.