Mark the tick against the correct answer in the following:

Question:

Mark the tick against the correct answer in the following:

If $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$ then $\left(\cos ^{-1} x+\cos ^{-1} y\right)=?$

A. $\frac{\pi}{6}$

B. $\frac{\pi}{3}$

C. $\pi$

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

 

Solution:

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

To Find: The value of $\cos ^{-1} x+\cos ^{-1} y$

Since we know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$

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

Similarly $\cos ^{-1} y=\frac{\pi}{2}-\sin ^{-1} y$

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

$=\frac{2 \pi}{2}-\left[\sin ^{-1} x+\sin ^{-1} y\right]$

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

$=\frac{\pi}{3}$

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