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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

The value of (2 ^ -1 x+ ^ -1 x) at x=(1)/(5), where 0 ≤ ^ -1 x ≤ π...

The value of cos⁡\cos\left(2 \cos ^{-1} x+\sin ^{-1} x\right)atx=15,whereat x=\frac{1}{5}, where0 \leq \cos ^{-1} x \leq \piand−π2≤sin⁡−1and -\frac{\pi}{2} \leq \sin ^{-1}x \leq \frac{\pi}{2}, is

Options

  1. A−65-\frac{\sqrt{6}}{5}
  2. B65\frac{\sqrt{6}}{5}
  3. C−265-\frac{2 \sqrt{6}}{5}
  4. D265\frac{2 \sqrt{6}}{5}

Correct answer

C. −265-\frac{2 \sqrt{6}}{5}

Step-by-step solution

cos⁡(2cos⁡−1x+sin⁡−1x)=cos⁡(cos⁡−1x+cos⁡−1x+sin⁡−1x)=cos⁡(cos⁡−1x+π2)[∵cos⁡−1x+sin⁡−1x=π2]=−sin⁡(cos⁡−1x)=−sin⁡(sin⁡−11−x2)=−1−x2=−1−(15)2[∵x=15]=−265\begin{aligned} & \cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)=\cos \left(\cos ^{-1} x+\cos ^{-1} x+\sin ^{-1} x\right) \\ & =\cos \left(\cos ^{-1} x+\frac{\pi}{2}\right) \quad\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}\right] \\ & =-\sin \left(\cos ^{-1} x\right) \\ & =-\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) \\ & =-\sqrt{1-x^2} \\ & =-\sqrt{1-\left(\frac{1}{5}\right)^2} \quad\left[\because x=\frac{1}{5}\right] \\ & =-\frac{2 \sqrt{6}}{5}\end{aligned}
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