JEE Main 2025 — Inverse Trigonometric Functions Question with Solution

$\begin{aligned} & \text { Let } \tan ^{-1} \alpha=A \Rightarrow \tan A=\alpha \\ & \cot ^{-1} \beta=B \Rightarrow \cot B=\beta \\ & \sec ^2 A+\operatorname{cosec}^2 B=36 \\ & \Rightarrow 1+\tan ^2 A+1+\cot ^2 B=36 \\ & \Rightarrow \alpha^2+\beta^2=34 \end{aligned}$ Also $\alpha +\beta =8$ (Given) $\begin{aligned} & \therefore(\alpha+\beta)^2=34+2 \alpha \beta=64 \\ & \Rightarrow \alpha \beta=15 \end{aligned}$ $\Rightarrow \alpha ,\beta$ are roots of equation $\begin{aligned} & x^2-8 x+15=0 \\ & \Rightarrow(x-3)(x-5)=0 \\ & \Rightarrow \quad x=3,5 \\ & \therefore \quad \alpha=3, \beta=5 \quad(\alpha < \beta) \\ & \therefore \quad \alpha^2+\beta=9+5=14 \end{aligned}$