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

 अत: ${\sin ^{ - 1}}(x\sqrt {1 - x} - \sqrt x \,\sqrt {1 - {x^2}} )$

$ = {\sin ^{ - 1}}(\sin \theta \sqrt {1 - {{\sin }^2}\phi  } - \sin \phi \sqrt {1 - {{\sin }^2}\theta } )$

$ = {\sin ^{ - 1}}(\sin \theta \cos \phi - \sin \phi \cos \theta ) = {\sin ^{ - 1}}\sin \,(\theta - \phi )$

$ = \theta - \phi = {\sin ^{ - 1}}(x) - {\sin ^{ - 1}}(\sqrt x )$.