Thank you for reporting, we will resolve it shortly
Q.
Value of $k$ for which the point $\left(\alpha, \sin ^{-1} \alpha\right)(\alpha>0)$ lies inside the triangle formed by $x+y=k$ with co-ordinate axes is -
Inverse Trigonometric Functions
Solution:
$\because \alpha>0 \Rightarrow\left(\alpha, \sin ^{-1} \alpha\right)$ lies in $1^{\text {st }}$ quandrant $\Rightarrow k >0$
Also the extreme point on the graph of $y =\sin ^{-1} x$ is
$\left(1, \frac{\pi}{2}\right)$
$\Rightarrow 1+\frac{\pi}{2}- k <0 $
$\Rightarrow k >1+\frac{\pi}{2}$
$\Rightarrow k \in\left(1+\frac{\pi}{2}, \infty\right)$