Answer $B$
Given:
$$\sin (\sin^{-1}\frac{2}{5} + \cos^{-1}x) = 1\qquad \to(1)$$
But we know that:
$$\sin \frac{\pi}{2} = 1$$
Substituting this value in in $(1)$, we get:
$$\sin (\sin^{-1}\frac{2}{5} + \cos^{-1}x) = \sin\frac{\pi}{2}$$
On comparing $\sin$ with $\sin$, we get:
$$\sin^{-1}\frac{2}{5} + \cos^{-1}x = \frac{\pi}{2}\qquad \to (2)$$
Also, we know that:
$$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\qquad \to (3)$$
So, on comparing $(2)$ and $(3)$, we get:
$$x = \frac{2}{5}$$
$\therefore \;B$ is the correct option.
