MHT CET · Maths · Inverse Trigonometric Functions
If \(\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1\), then the value of \(x\) is
- A \(\frac{\pi}{2}+\frac{1}{5}\)
- B \(\frac{\pi}{2}-\frac{1}{5}\)
- C \(-\frac{1}{5}\)
- D \(\frac{1}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{5}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin \sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1=\sin \frac{\pi}{2} \\ & \Rightarrow \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2} \\ & \Rightarrow \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} \frac{1}{5} \\ & \Rightarrow \cos ^{-1} x=\cos ^{-1} \frac{1}{5} \\ & \Rightarrow x=\frac{1}{5}\end{aligned}\)
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