Question

If the minimum value of ${\left({\sec }^{-1}x\right)}^2+{\left(cose{c}^{-1}x\right)}^2$ is $\frac{{\pi }^a}{b}$. Find the value of $a+b$

• A. $6$
• B. $8$
• C. $10$
• D. $12$

Answer 1

The correct option is C $10$

Let, $I={\left({\sec }^{-1}x\right)}^2+{\left(cose{c}^{-1}x\right)}^2$

$={({\sec }^{-1}x+cose{c}^{-1}x)}^2-2{\sec }^{-1}x.cose{c}^{-1}x$, Using Identity $[{\sec }^{-1}x+cose{c}^{-1}x=\frac{\pi }{2}]$

$=\frac{{\pi }^2}{4}-2{\sec }^{-1}x(\frac{\pi }{2}-{\sec }^{-1}x)$

$=\frac{{\pi }^2}{4}+2{\left({\sec }^{-1}x\right)}^2-\pi {\sec }^{-1}x$

$=\frac{{\pi }^2}{4}+2[{\left({\sec }^{-1}x\right)}^2-2\frac{\pi }{4}{\sec }^{-1}x-\frac{{\pi }^2}{8}]$

$=2{({\sec }^{-1}x-\frac{\pi }{4})}^2+\frac{{\pi }^2}{8}$ $\Rightarrow$ $I\ge \frac{{\pi }^2}{8}$