Question

The maximum value of $(se{c}^{-1}x{)}^2+(cose{c}^{-1}x{)}^2$ is equal to

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

Answer 1

The correct option is B

$\frac{5{\pi }^2}{4}$

$I=(se{c}^{-1}x{)}^2+(cose{c}^{-1}x{)}^2$
$=(se{c}^{-1}x+cose{c}^{-1}x{)}^2-2se{c}^{-1}xcose{c}^{-1}x$
$=\frac{{\pi }^2}{4}-2se{c}^{-1}x(\frac{\pi }{2}-se{c}^{-1}x)$
$=\frac{{\pi }^2}{4}+2\left(\right(se{c}^{-1}x{)}^2-\frac{\pi }{2}\left(se{c}^{-1}x\right)+\frac{{\pi }^2}{16}-\frac{{\pi }^2}{16})$
$=\frac{{\pi }^2}{4}+2{(se{c}^{-1}x-\frac{\pi }{4})}^2-\frac{{\pi }^2}{8}\Rightarrow I_{max}=\frac{{\pi }^2}{4}+2\times \frac{9{\pi }^2}{16}-\frac{{\pi }^2}{8}$
$={\frac{5\pi }{4}}^2$