Maths Q6 – JEE-MAIN 2026

Home Maths Complex & QuadraticQuestion #77

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:

Clue 1 of 3: Roots of Quadratic Equation

Use Vieta's formulas to find the sum and product of the roots, $\tan A + \tan B$ and $\tan A \tan B$ .

🧠 Full Solution Path

Step 1: Find sum and product of roots and

\tan (A + B)

✦ Active

The roots of $x^{2} - 2 x - 5 = 0$ are $\tan A$ and $\tan B$ . Using Vieta's formulas:

\tan A + \tan B = 2

\tan A \tan B = - 5

Now, calculate $\tan (A + B)$ using the tangent addition formula:

\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (- 5)} = \frac{2}{6} = \frac{1}{3}

Step 2: Determine

\cos (A + B)

and its sign○ Expand

From $\tan (A + B) = \frac{1}{3}$ , we can construct a right triangle with opposite side 1 and adjacent side 3. The hypotenuse is $\sqrt{1^{2} + 3^{2}} = \sqrt{10}$ .

The roots of the quadratic equation are $x = \frac{- (- 2) \pm \sqrt{(- 2)^{2} - 4 (1) (- 5)}}{2 (1)} = \frac{2 \pm \sqrt{4 + 20}}{2} = \frac{2 \pm \sqrt{24}}{2} = 1 \pm \sqrt{6}$ . So, $\tan A = 1 + \sqrt{6}$ and $\tan B = 1 - \sqrt{6}$ .

Since $A, B \in (- π / 2, π / 2)$ , and $\tan A = 1 + \sqrt{6} > 0$ , $A \in (0, π / 2)$ . Since $\tan B = 1 - \sqrt{6} < 0$ , $B \in (- π / 2, 0)$ . Also, $1 + \sqrt{6} > 1 ⟹ A > π / 4$ . And $1 - \sqrt{6} < - 1 ⟹ B < - π / 4$ . Therefore, $A + B \in (- π / 4, π / 4)$ . In this interval, $\cos (A + B)$ is positive.

Thus, $\cos (A + B) = \frac{adjacent}{hypotenuse} = \frac{3}{\sqrt{10}}$ .

Step 3: Calculate

20 \sin^{2} (\frac{A + B}{2})

○ Expand

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

\sin^{2} (\frac{A + B}{2}) = \frac{1 - \cos (A + B)}{2} = \frac{1 - \frac{3}{\sqrt{10}}}{2} = \frac{\sqrt{10} - 3}{2 \sqrt{10}}

Finally, calculate the required expression:

20 \sin^{2} (\frac{A + B}{2}) = 20 \times \frac{\sqrt{10} - 3}{2 \sqrt{10}} = \frac{10 (\sqrt{10} - 3)}{\sqrt{10}}

= \frac{10 \sqrt{10} - 30}{\sqrt{10}} = \frac{(10 \sqrt{10} - 30) \sqrt{10}}{10} = \frac{10 (10) - 30 \sqrt{10}}{10} = \frac{100 - 30 \sqrt{10}}{10} = 10 - 3 \sqrt{10}

💡 Teacher's Secret Hint

Remember to rationalize the denominator for the final simplification.

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Maths Question 6 – JEE-MAIN 2026

Clue 1 of 3: Roots of Quadratic Equation

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