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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

BITSAT 2023 · Paper 2 · 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. D
    None 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+π/2]=−sin⁡[cos⁡−1x]=−sin⁡[sin⁡−11−x2]=−1−x2=−1−(15)2=−2425=−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+\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} \\ & \end{aligned}
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