If $\sin^{-1} x = \frac{\pi}{5}$ for some $x \in (-1, 1)$,then the value of $\cos^{-1} x$ is

  • A
    $\frac{3\pi}{10}$
  • B
    $\frac{5\pi}{10}$
  • C
    $\frac{7\pi}{10}$
  • D
    $\frac{9\pi}{10}$

Explore More

Similar Questions

Find the value of $\tan \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2} \right]$,where $|x|  <1, y>0$ and $xy < 1$.

If $\cos^{-1} \frac{3}{5} - \sin^{-1} \frac{4}{5} = \cos^{-1} x$,then $x = $

The solution of $\sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3}$ is

If $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$, then $\cos(2 \csc^{-1}\sqrt{k - 1}) = \dots$

If ${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x$,then $x = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo