Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

If $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$, then the value of $\cos ^{-1} x+\cos ^{-1} y$ is :

Options:

$\frac{\pi}{2}$

$\frac{\pi}{3}$

$\frac{2 \pi}{3}$

$\frac{\pi}{6}$

Correct Answer:

$\frac{\pi}{3}$

Explanation:

$\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$

We know that

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

so  $\cos -1 x=\frac{\pi}{2}^{-1}-\sin ^{-1} x $

or  $\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x$

$\frac{\pi}{2}-\cos ^{-1} x+\frac{\pi}{2}-\cos ^{-1} y=\frac{2 \pi}{3}$

so $\pi-\left(\cos ^{-1} x \cos ^{-1} y\right)=\frac{2 \pi}{3}$

so  $\pi-\frac{2 \pi}{3}=\cos ^{-1} x+\cos ^{-1} y$

so  $\frac{\pi}{3}=\cos ^{-1} x + \cos^{-1} y$