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  4. The maximum value of( sec -1 x)2+(cosec -1 x)2 is equal to

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

Inverse Trigonometric Functions

Solution:

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