ISI2016-DCG-66

Question

If $\sin(\sin^{-1}\frac{2}{5}+\cos^{-1}x)=1,$ then $x$ is

• A. $1$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. None of these

Answer 1

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.