sec²(tan⁻¹α)+cosec²(cot⁻¹β)=36, α+β=8, α<β — Find α²+β

JEE Main 2025 — 24 January, Morning Shift. Previous Year Question.

Problem

If $\alpha$ and $\beta$ are real numbers such that $\sec^2(\tan^{-1}\alpha)+\text{cosec}^2(\cot^{-1}\beta)=36$ and $\alpha+\beta=8$ (with $\alpha<\beta$), then $\alpha^2+\beta$ is:

Key insight. $\sec^2(\tan^{-1}\alpha)$ looks intimidating, but $\tan^{-1}\alpha$ is just “the angle whose tangent is $\alpha$” — so $\tan(\tan^{-1}\alpha)=\alpha$ directly, and $\sec^2\theta=1+\tan^2\theta$ turns the whole expression into $1+\alpha^2$. The same trick applies to the cosec/cot pair, collapsing a scary-looking trigonometric equation into plain algebra in $\alpha$ and $\beta$.

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Approach

Simplify both inverse-trig expressions using the identities $\sec^2\theta=1+\tan^2\theta$ and $\text{cosec}^2\theta=1+\cot^2\theta$, turning the given equation into $\alpha^2+\beta^2=34$. Combine this with $\alpha+\beta=8$ using the standard algebraic identities linking sums, products, and differences of two numbers, to solve for $\alpha$ and $\beta$ individually.

Solution

Step 1 — Simplify sec²(tan⁻¹α)

$$\sec^2(\tan^{-1}\alpha) = 1+\tan^2(\tan^{-1}\alpha) = 1+\alpha^2$$

Step 2 — Simplify cosec²(cot⁻¹β)

$$\text{cosec}^2(\cot^{-1}\beta) = 1+\cot^2(\cot^{-1}\beta) = 1+\beta^2$$

Step 3 — Substitute into the given equation

$$(1+\alpha^2)+(1+\beta^2) = 36 \implies \alpha^2+\beta^2 = 34$$

Step 4 — Find αβ using the given sum

$$(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta \implies 64 = 34+2\alpha\beta \implies \alpha\beta = 15$$

Step 5 — Find α−β

$$(\alpha-\beta)^2 = (\alpha+\beta)^2-4\alpha\beta = 64-60 = 4 \implies \alpha-\beta = \pm2$$

Since $\alpha<\beta$, $\alpha-\beta$ must be negative:

$$\alpha-\beta = -2$$

Step 6 — Solve for α and β

Adding $\alpha+\beta=8$ and $\alpha-\beta=-2$:

$$2\alpha = 6 \implies \alpha=3, \qquad \beta = 8-3 = 5$$

Step 7 — Compute α²+β

$$\alpha^2+\beta = 9+5 = 14$$

Answer

Common mistakes

• Picking the wrong sign for α−β. Both $\alpha-\beta=2$ and $\alpha-\beta=-2$ satisfy the squared equation, but only $-2$ is consistent with the given condition $\alpha<\beta$ — choosing the wrong sign swaps the values of $\alpha$ and $\beta$ and gives a different (wrong) final answer.
• Forgetting the “$+1$” when converting sec² and cosec². It’s easy to write $\sec^2(\tan^{-1}\alpha)=\alpha^2$ directly, skipping the identity $\sec^2\theta=1+\tan^2\theta$ — this drops a constant term that changes the entire equation.

Practise next

• If $\gamma,\delta$ are real numbers with $\sec^2(\tan^{-1}\gamma)+\operatorname{cosec}^2(\cot^{-1}\delta)=60$ and $\gamma+\delta=10$, $\gamma>\delta$, find $\gamma+\delta^2$ using the same approach.