Theory of EquationsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

20 sin²((A+B)/2) = 10 - 3√10 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let tan⁡A,tan⁡B\tan A,\tan B, where A,B∈(−π2,π2)A,B\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right), be the roots of x2−2x−5=0x^2-2x-5=0. Then 20sin⁡2(A+B2)20\sin^2\left(\dfrac{A+B}{2}\right) is equal to
A10+1010+\sqrt{10}
B10−21010-2\sqrt{10}
C10−31010-3\sqrt{10}correct
D10−1010-\sqrt{10}
Solution
Step 1: By Vieta on x2−2x−5=0x^2-2x-5=0:
tan⁡A+tan⁡B=2,tan⁡Atan⁡B=−5.\tan A+\tan B=2,\qquad \tan A\tan B=-5.
 Step 2: tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B=21−(−5)=26=13\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{2}{1-(-5)}=\dfrac{2}{6}=\dfrac13.
∴ cos⁡(A+B)=312+32=310.\therefore\ \cos(A+B)=\frac{3}{\sqrt{1^2+3^2}}=\frac{3}{\sqrt{10}}.
 Step 3: sin⁡2 ⁣(A+B2)=1−cos⁡(A+B)2\sin^2\!\left(\dfrac{A+B}{2}\right)=\dfrac{1-\cos(A+B)}{2}.
20sin⁡2 ⁣(A+B2)=10(1−cos⁡(A+B))=10(1−310).20\sin^2\!\left(\frac{A+B}{2}\right)=10\big(1-\cos(A+B)\big)=10\left(1-\frac{3}{\sqrt{10}}\right).
 Step 4: 1010=10\dfrac{10}{\sqrt{10}}=\sqrt{10}:
10−3010=10−310.10-\frac{30}{\sqrt{10}}=10-3\sqrt{10}.
 Correct answer: (3)
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