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

The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$ is
  1. $-\frac{\sqrt{6}}{5}$
  2. $\frac{2 \sqrt{6}}{5}$
  3. $-\frac{2 \sqrt{6}}{5}$
  4. $\frac{2 \sqrt{5}}{6}$

Solution

$\begin{aligned} & \cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right) \\ & =\cos \left[\left(\sin ^{-1} x+\cos ^{-1} x\right)+\cos ^{-1} x\right] \\ & =\cos \left(\frac{\pi}{2}+\cos ^{-1} x\right) \\ & =-\sin \left(\cos ^{-1} x\right) \\ & =-\sin \left(\sin ^{-1} \sqrt{\left(1-x^2\right)}\right) \\ & =-\sqrt{1-x^2} \\ & =-\sqrt{1-\left(\frac{1}{5}\right)^2} \\ & =-\sqrt{\frac{24}{25}} \\ & =-\frac{2 \sqrt{6}}{5}\end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

Practice more Inverse Trigonometric Functions questions on Aicharya