Inverse Trigonometric Functions — Class 12 Maths Solution

$\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1$
Or ${\sin ^{ - 1}}\frac{1}{5} + {\cos ^{ - 1}}x = {\sin ^{ - 1}}1$

$\Rightarrow$ ${\sin ^{ - 1}}\frac{1}{5} + \frac{\pi }{2} - {\sin ^{ - 1}}x = \frac{\pi }{2} \Rightarrow {\sin ^{ - 1}}\frac{1}{5} = {\sin ^{ - 1}}x$

$\Rightarrow$ $x = \sin \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right) = \frac{1}{5}.$