JEE Main 2025 — Mathematics Trigonometry

$$\begin{aligned} & 16\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2 \\ & \operatorname{Sec}^{-1} x=a \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \\ & \operatorname{cosec}^{-1} x=\frac{\pi}{2}-a \\ & =16\left[a^2+\left(\frac{\pi}{2}-a\right)^2\right]=16\left[2 a^2-\pi a+\frac{\pi^2}{4}\right] \\ & \max ]_{a=\pi}=16\left[2 \pi^2-\pi^2+\pi \frac{2}{4}\right]=20 \pi^2 \\ & \min ]_{a=\frac{\pi}{4}}=16\left[\frac{2 \times \pi^2}{16}-\frac{\pi^2}{4}+\frac{\pi^2}{4}\right]=2 \pi^2 \\ & \operatorname{Sum}=22 \pi^2\end{aligned}$$​16(sec−1x)2+(cosec−1x)2Sec−1x=a∈[0,π]−{2π​}cosec−1x=2π​−a=16[a2+(2π​−a)2]=16[2a2−πa+4π2​]max]a=π​=16[2π2−π2+π42​]=20π2min]a=4π​​=16[162×π2​−4π2​+4π2​]=2π2Sum=22π2​