Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & 16\left(\sec ^{-1} xight)^2+\left(\operatorname{cosec}^{-1} xight)^2 \ & \operatorname{sec}^{-1} x=a \in[0, \pi]-\left\{\frac{\pi}{2}ight\} \ & \operatorname{cosec}^{-1} x=\frac{\pi}{2}-a \ & =16\left[a^2+\left(\frac{\pi}{2}-aight)^2ight]=16\left[2 a^2-\pi a+\frac{\pi^2}{4}ight] \ & \max ]_{a=\pi}=16\left[2 \pi^2-\pi^2+\pi \frac{2}{4}ight]=20 \pi^2 \ & \min ]_{a=\frac{\pi}{4}}=16\left[\frac{2 	imes \pi^2}{16}-\frac{\pi^2}{4}+\frac{\pi^2}{4}ight]=2 \pi^2 \ & \operatorname{Sum}=22 \pi^2\end{aligned}$

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum…

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