CUET Preparation Today
If $\sin^{-1}x=\frac{π}{5}$ for some x ∈ [-1, 1], then the value of $\cos^{-1} x$ is: |
$\frac{9π}{10}$ $\frac{7π}{10}$ $\frac{5π}{10}$ $\frac{3π}{10}$ |
$\frac{3π}{10}$ |
$\sin^{-1}x+\cos^{-1}=\frac{π}{2}⇒\cos^{-1}=\frac{π}{2}-\sin^{-1}x=\frac{π}{2}-\frac{π}{5}=\frac{3π}{10}$ |
