Question

The maximum value of${\left({\sec }^{-1}x\right)}^2+{\left(cosec{}^{-1}x\right)}^2$ is equal to

• A. $\frac{{\pi }^2}{2}$
• B. $\frac{5{\pi }^2}{4}$
• C. ${\pi }^2$
• D. None of these

Answer 1

Answer: (B)

$I={\left({\sec }^{-1}x\right)}^2+{\left(cosec{}^{-1}x\right)}^2$
$={\left({\sec }^{-1}x+cosec{}^{-1}x\right)}^2-2{\sec }^{-1}x\cdot cosec{}^{-1}x$
$=\frac{{\pi }^2}{4}-2{\sec }^{-1}x\left(\frac{\pi }{2}-{\sec }^{-1}x\right)$
$=\frac{{\pi }^2}{4}+2\left({\left({\sec }^{-1}x\right)}^2-\frac{\pi }{2}\cdot \left({\sec }^{-1}x\right)+\frac{{\pi }^2}{16}-\frac{{\pi }^2}{16}\right)$
$\Rightarrow \frac{{\pi }^2}{4}+2\left[{\left({\sec }^{-1}x-\frac{\pi }{4}\right)}^2\right]$
$\Rightarrow I_{\max }=\frac{{\pi }^2}{18}+2\left[\frac{9{\pi }^2}{16}\right]$
$\Rightarrow \frac{{\pi }^2}{8}+\frac{9{\pi }^2}{8}$
$\Rightarrow I_{\max }=\frac{5{\pi }^2}{4}$