JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Inverse Trigonometric Functions

For $|x|\ge 1$, ${\sec }^{-1}x+{\mathrm{cosec}}^{-1}x=\frac{\pi }{2}$ always. Write ${\sec }^{-1}x=a\in [0,\pi ]\setminus \{\pi /2\}$, so ${\mathrm{cosec}}^{-1}x=\frac{\pi }{2}-a$.

The expression becomes


This is an upward parabola in $a$, so its minimum over the interval occurs at the vertex $a=\pi /4$ (a valid value, corresponding to $x=\sqrt{2}$):


Being an upward parabola on the closed interval $[0,\pi ]$, its maximum occurs at an endpoint. Both $a=0$ (i.e. $x=1$) and $a=\pi$ (i.e. $x=-1$) are attainable, and

The larger value is $20{\pi }^2$, at $a=\pi$.

Sum of maximum and minimum: