JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Two adjacent sides of a parallelogram $P QR S$ are given by $P Q = \hat{j} + \hat{k}$ and $P S = \hat{i} - \hat{j}$ . If the side $P S$ is rotated about the point $P$ by an acute angle $α$ in the plane of the parallelogram so that it becomes perpendicular to the side $P Q$ , then $sin^{2} (\frac{5 α}{2}) - sin^{2} (\frac{α}{2})$ is equal to :

A
$\frac{1}{2}$
B
$\frac{3}{2}$
C
$\frac{3}{4}$
D
$\frac{2 3}{5}$

Solution

The angle $θ$ between $P Q$ and $P S$ satisfies $cos θ = \frac{P Q \cdot P S}{∣ P Q ∣∣ P S ∣} = \frac{- 1}{2}$ , so $θ = \frac{2 π}{3}$ .

Rotating $P S$ by $α$ to become perpendicular to $P Q$ means $θ - α = \frac{π}{2}$ , hence $α = \frac{2 π}{3} - \frac{π}{2} = \frac{π}{6}$ .

Using $sin^{2} A - sin^{2} B = sin (A + B) sin (A - B)$ with $A = \frac{5 α}{2}$ and $B = \frac{α}{2}$ :

$sin^{2} \frac{5 α}{2} - sin^{2} \frac{α}{2} = sin 3 α \cdot sin 2 α = sin \frac{π}{2} \cdot sin \frac{π}{3} = 1 \cdot \frac{3}{2} = \frac{3}{2}$ .

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