If $sin \left( sin^{-1} \dfrac{1}{5} + cos^{-1} x \right) = 1$, find $x$

The correct answer is $\dfrac{1}{5}$

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


⇒ $\left( sin^{-1} \dfrac{1}{5} + cos^{-1} x \right) $ = $sin^{-1}(1)$ = $\dfrac{\pi}{2}$


∴ $ sin^{-1} \dfrac{1}{5} + cos^{-1} x $ = $\dfrac{\pi}{2}$


We know that $sin^{-1} x + cos^{-1} x = \dfrac{\pi}{2}$


∴ $x = \dfrac{1}{5} $