Let alpha=3+4+8+9+13+14+... upto 40 terms. If (tanbeta)^(alpha/1020) is a root of the equation x^2+x-2=0, beta in (0, pi/2), then sin^2beta+3cos^2beta is equal to :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q5 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #401

Let

α = 3 + 4 + 8 + 9 + 13 + 14 + . . .

upto 40 terms. If

(\tan β)^{\frac{α}{1020}}

is a root of the equation

x^{2} + x - 2 = 0

β \in (0, \frac{π}{2})

, then

\sin^{2} β + 3 \cos^{2} β

is equal to :

Clue 1 of 3: Series Summation

Group the terms of the series for $α$ into pairs to identify an arithmetic progression and calculate its sum.

🧠 Full Solution Path

Step 1: Calculate the value of

α

✦ Active

The given series for $α$ is $3 + 4 + 8 + 9 + 13 + 14 + . . .$ upto 40 terms. We can group the terms in pairs:

α = (3 + 4) + (8 + 9) + (13 + 14) + . . . (20 pairs)

This simplifies to an arithmetic progression (AP) of sums:

α = 7 + 17 + 27 + . . . (20 terms)

For this AP, the first term is $A = 7$ , the common difference is $D = 10$ , and the number of terms is $N = 20$ . The sum is given by $S_{N} = \frac{N}{2} [2 A + (N - 1) D]$ :

α = \frac{20}{2} [2 (7) + (20 - 1) 10] = 10 [14 + 190] = 10 [204] = 2040

Step 2: Solve the quadratic equation and find

\tan β

○ Expand

The given quadratic equation is $x^{2} + x - 2 = 0$ . Factoring this equation gives:

(x + 2) (x - 1) = 0

The roots are $x = - 2$ and $x = 1$ . We are given that $(\tan β)^{\frac{α}{1020}}$ is a root. Substitute $α = 2040$ :

(\tan β)^{\frac{2040}{1020}} = (\tan β)^{2} = \tan^{2} β

Since $β \in (0, \frac{π}{2})$ , $\tan β$ must be positive, which means $\tan^{2} β$ must also be positive. Therefore, $\tan^{2} β$ must be equal to the positive root of the quadratic equation:

\tan^{2} β = 1

Since $\tan β > 0$ for $β \in (0, \frac{π}{2})$ , we have:

\tan β = 1

This implies $β = \frac{π}{4}$ .

💡 Teacher's Secret Hint

Remember to consider the domain of $β$ when choosing the root for $\tan β$ .

Step 3: Evaluate the final expression○ Expand

We need to find the value of $\sin^{2} β + 3 \cos^{2} β$ . Substitute $β = \frac{π}{4}$ :

\sin^{2} (\frac{π}{4}) + 3 \cos^{2} (\frac{π}{4})

We know that $\sin (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ and $\cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ .

{(\frac{1}{\sqrt{2}})}^{2} + 3 {(\frac{1}{\sqrt{2}})}^{2} = \frac{1}{2} + 3 (\frac{1}{2}) = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2

💡 Teacher's Secret Hint

Alternatively, use the identity $\sin^{2} β + \cos^{2} β = 1$ to simplify the expression to $1 + 2 \cos^{2} β$ before substituting $β$ .

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Maths Question 5 – JEE-MAIN 2026

Clue 1 of 3: Series Summation

🧠 Full Solution Path