If $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$, then the value of $\cos ^{-1} x+\sin ^{-1} y$ is
If $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$, then the value of $\cos ^{-1} x+\sin ^{-1} y$ is
- $\frac{\pi}{10}$
- $\frac{7\pi}{10}$
- $\frac{9\pi}{10}$
- $\frac{3\pi}{10}$
Solution
$\begin{array}{ll} & \sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \frac{\pi}{2}-\cos ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \pi-\cos ^{-1} x-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \cos ^{-1} x+\sin ^{-1} y=\pi-\frac{3 \pi}{10}=\frac{7 \pi}{10}\end{array}$
Asked in: MHT CET 2023 (09 May Shift 1)
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