JEE Main 2026 Apr 5 Shift 1, Mathematics Q6: Quadratic Equations

JEE Main2026Apr 5, Shift 1Mathematics

Let $tan A, tan B$ , where $A, B \in (- \frac{π}{2}, \frac{π}{2})$ , be the roots of the quadratic equation $x^{2} - 2 x - 5 = 0$ . Then $20 sin^{2} (\frac{A + B}{2})$ is equal to :

A
$10 + 10$
B
$10 - 210$
C
$10 - 310$
D
$10 - 10$

Solution

By Vieta's formulas, $tan A + tan B = 2$ and $tan A tan B = - 5$ .

Then $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B} = \frac{2}{1 - ( - 5 )} = \frac{1}{3}$ , so $cos (A + B) = \frac{3}{10}$ .

Using the half-angle identity $2 sin^{2} θ = 1 - cos 2 θ$ :

$20 sin^{2} (\frac{A + B}{2}) = 10 (1 - cos (A + B)) = 10 (1 - \frac{3}{10}) = 10 - 310$ .

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JEE Main 2026 Apr 5 Shift 1, Mathematics Q6: Quadratic Equations

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