$\begin{aligned}
& \text { Let } \tan ^{-1} \alpha=A \Rightarrow \tan A=\alpha \
& \cot ^{-1} \beta=B \Rightarrow \cot B=\beta \
& \sec ^2 A+\operatorname{cosec}^2 B=36 \
& \Rightarrow 1+\tan ^2 A+1+\cot ^2 B=36 \
& \Rightarrow \alpha^2+\beta^2=34
\end{aligned}Also\alpha+\beta=8(Given)\begin{aligned}
& \therefore(\alpha+\beta)^2=34+2 \alpha \beta=64 \
& \Rightarrow \alpha \beta=15
\end{aligned}\Rightarrow \alpha, \betaarerootsofequation\begin{aligned}
& x^2-8 x+15=0 \
& \Rightarrow(x-3)(x-5)=0 \
& \Rightarrow \quad x=3,5 \
& \therefore \quad \alpha=3, \beta=5 \quad(\alpha < \beta) \
& \therefore \quad \alpha^2+\beta=9+5=14
\end{aligned}$
If for some α, β ; α ≤ β, α+β-8 and sec ^2 (tan ^-1 α )+ cosec^2 (cot ^-1 β )-36, then α^2+β is.
JEE Main 2025 — Mathematics Trigonometry
If for some α,β;α≤β,α+β−8 and sec2(tan−1α)+cosec2(cot−1β)−36, then α2+β is_______.
Held on 24 Jan 2025 · Verified 6 Jul 2026.
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