MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)\) at \(x=\frac{1}{5}\) is
- A \(-\frac{\sqrt{6}}{5}\)
- B \(\frac{2 \sqrt{6}}{5}\)
- C \(-\frac{2 \sqrt{6}}{5}\)
- D \(\frac{2 \sqrt{5}}{6}\)
Answer & Solution
Correct Answer
(C) \(-\frac{2 \sqrt{6}}{5}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right) \\ & =\cos \left[\left(\sin ^{-1} x+\cos ^{-1} x\right)+\cos ^{-1} x\right] \\ & =\cos \left(\frac{\pi}{2}+\cos ^{-1} x\right) \\ & =-\sin \left(\cos ^{-1} x\right) \\ & =-\sin \left(\sin ^{-1} \sqrt{\left(1-x^2\right)}\right) \\ & =-\sqrt{1-x^2} \\ & =-\sqrt{1-\left(\frac{1}{5}\right)^2} \\ & =-\sqrt{\frac{24}{25}} \\ & =-\frac{2 \sqrt{6}}{5}\end{aligned}\)
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