CUET Preparation Today
The set of value sof x satisfying $|sin^{-1}x|<|cos^{-1}x|$, is |
$[-1, 1/\sqrt{2}]$ $[-1, 1/\sqrt{2}] ∪ [1/\sqrt{2},1]$ $(-1, 1/\sqrt{2})$ none of these |
$[-1, 1/\sqrt{2}]$ |
We know that $0 ≤ cos^{-1} x ≤ \pi $. $ ∴ |cos^{-1}x|= cos^{-1} x$
It is evident from the graphs of $y = |sin^{-1}x|$ and $y = |cos^{-1}x|$ that $|sin^{-1}x| < |cos^{-1}x|$ for all x ∈$ [-1, 1/\sqrt{2}]$ |
