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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

BITSAT 2021 · Mathematics · Inverse Trigonometric Functions

The value of (2 ^ -1 x+ ^ -1 x) at x=(1)/(5)is

The value of cos⁡\cos\left(2 \cos ^{-1} x+\sin ^{-1} x\right)atatx=\frac{1}{5}is

Options

  1. A−265-\frac{2 \sqrt{6}}{5}
  2. B−26-2 \sqrt{6}
  3. C−65-\frac{\sqrt{6}}{5}
  4. DNone of these

Correct answer

A. −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]=−sin⁡[cos⁡−1x]=−sin⁡[sin⁡−11−x2]=−1−x2=−1−(15)2=−2425=−265\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]=-\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}}=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5}
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