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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Inverse Trigonometric Functions

If for some α, β ; α ≤ β, α+β-8 and ^2( ^ -1 α)+cosec^2( ^ -1 β)-36...

If for some α,β;α≤β,α+β−8\alpha, \beta ; \alpha \leq \beta, \alpha+\beta-8 and sec⁡2(tan⁡−1α)+cosec⁡2(cot⁡−1β)−36\sec ^2\left(\tan ^{-1} \alpha\right)+\operatorname{cosec}^2\left(\cot ^{-1} \beta\right)-36, then α2+β\alpha^2+\beta is_______.

Correct answer

0

Step-by-step solution

 Let tan⁡−1α=A⇒tan⁡A=αcot⁡−1β=B⇒cot⁡B=βsec⁡2A+cosec⁡2B=36⇒1+tan⁡2A+1+cot⁡2B=36⇒α2+β2=34\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 α+β=8\alpha+\beta=8 (Given) ∴(α+β)2=34+2αβ=64⇒αβ=15\begin{aligned} & \therefore(\alpha+\beta)^2=34+2 \alpha \beta=64 \\ & \Rightarrow \alpha \beta=15 \end{aligned}\Rightarrow \alpha, \betaarerootsofequationare roots of equationx2−8x+15=0⇒(x−3)(x−5)=0⇒x=3,5∴α=3,β=5(α<β)∴α2+β=9+5=14\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}$
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