MHT CET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}\), then the value of \(\cos ^{-1} x+\sin ^{-1} y\) is
- A \(\frac{\pi}{10}\)
- B \(\frac{7\pi}{10}\)
- C \(\frac{9\pi}{10}\)
- D \(\frac{3\pi}{10}\)
Answer & Solution
Correct Answer
(B) \(\frac{7\pi}{10}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \frac{\pi}{2}-\cos ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \pi-\cos ^{-1} x-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \cos ^{-1} x+\sin ^{-1} y=\pi-\frac{3 \pi}{10}=\frac{7 \pi}{10}\end{array}\)
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