If $\cos \left( \sin^{-1} \frac{2}{5} + \cos^{-1} x \right) = 0$, then $x$ is equal to

$\frac{2}{5}$

The correct answer is Option (2) → $\frac{2}{5}$ ##

We have, $\cos \left( \sin^{-1} \frac{2}{5} + \cos^{-1} x \right) = 0 \dots(i)$

Let $\sin^{-1} \frac{2}{5} = \alpha ⇒\sin \alpha = \frac{2}{5}$

$∴\cos \alpha = \sqrt{1 - \left( \frac{2}{5} \right)^2} = \frac{\sqrt{21}}{5}$

Again, let $\cos^{-1} x = \beta ⇒\cos \beta = x$

$∴\sin \beta = \sqrt{1 - x^2}$

From Eq. (i), we get

$\cos \left( \sin^{-1} \frac{2}{5} + \cos^{-1} x \right) = 0$

$⇒\cos (\alpha + \beta) = 0$

$⇒\cos \alpha \cos \beta - \sin \alpha \sin \beta = 0$

$⇒\frac{\sqrt{21}}{5} \cdot x - \frac{2}{5} \sqrt{1 - x^2} = 0$

$⇒x\sqrt{21} - 2\sqrt{1 - x^2} = 0$

$⇒x\sqrt{21} = 2\sqrt{1 - x^2}$

On squaring both sides, we get

$21x^2 = 4(1 - x^2)$

$⇒21x^2 = 4 - 4x^2$

$⇒25x^2 = 4 ⇒x^2 = \frac{4}{25}$

$∴x = \pm \frac{2}{5}$