JEE Main 2025 Jan 24 Shift 1, Mathematics Q22: Inverse Trigonometric Functions

JEE Main2025Jan 24, Shift 1Mathematics

If for some $α, β$ ; $α \leq β$ , $α + β = 8$ and $sec^{2} (tan^{- 1} α) + csc^{2} (cot^{- 1} β) = 36$ , then $α^{2} + β$ is ______.

Your answer

Solution

Use the standard identities $sec^{2} (tan^{- 1} α) = 1 + α^{2}$ and $csc^{2} (cot^{- 1} β) = 1 + β^{2}$ . The given relation becomes

$1 + α^{2} + 1 + β^{2} = 36 \Rightarrow α^{2} + β^{2} = 34$

Combined with $α + β = 8$ :

$(α + β)^{2} = α^{2} + β^{2} + 2 α β \Rightarrow 64 = 34 + 2 α β \Rightarrow α β = 15$

So $α, β$ are roots of $t^{2} - 8 t + 15 = 0$ , giving $t = 3, 5$ . With $α \leq β$ , $α = 3$ , $β = 5$ .

$α^{2} + β = 9 + 5 = 14$ .

$∴$ Answer: $14$ .

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