Inverse Trigonometric Functions – JEE Main 24 January 2025 Shift 1

Solution:

Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$ and $\text{cosec}^2 \theta = 1 + \cot^2 \theta$:
$$ \sec^2(\tan^{-1}\alpha) = 1 + [\tan(\tan^{-1}\alpha)]^2 = 1 + \alpha^2 $$
$$ \text{cosec}^2(\cot^{-1}\beta) = 1 + [\cot(\cot^{-1}\beta)]^2 = 1 + \beta^2 $$

Substitute these into the given equation:
$$ (1 + \alpha^2) + (1 + \beta^2) = 36 $$
$$ \alpha^2 + \beta^2 = 34 $$

We have the system:
1) $\alpha + \beta = 8$
2) $\alpha^2 + \beta^2 = 34$

Using $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$:
$$ (8)^2 = 34 + 2\alpha\beta $$
$$ 64 = 34 + 2\alpha\beta \Rightarrow 2\alpha\beta = 30 \Rightarrow \alpha\beta = 15 $$
The values $\alpha$ and $\beta$ are roots of $t^2 – 8t + 15 = 0$.
$$ (t-3)(t-5) = 0 \Rightarrow t = 3, 5 $$
Given $\alpha \le \beta$, we get $\alpha = 3$ and $\beta = 5$.

Calculate $\alpha^2 + \beta$:
$$ \alpha^2 + \beta = (3)^2 + 5 = 9 + 5 = 14 $$

Ans. (14)