CUET Preparation Today
If $cos^{-1} x > sin^{-1} x$, then x belong to the internal |
(-∞ , 0) (-1, 0) $[0, \frac{1}{\sqrt{2}})$ $[-1, \frac{1}{\sqrt{2}})$ |
$[-1, \frac{1}{\sqrt{2}})$ |
We know that $ cos^{-1}x$ and $sin^{-1}x$ exist for x ∈ [-1, 1]. Now, $cos^{-1}x > sin^{-1}x $
$⇒ cos^{-1}x> \frac{\pi}{2}- cos^{-1}x$ $⇒ 2cos^{-1} x > \frac{\pi}{2}⇒cos^{-1} x > \frac{\pi}{4} ⇒ x ∈ [-1, \frac{1}{\sqrt{2}})$ |
