Let $$\alpha = 3 + 4 + 8 + 9 + 13 + 14 + ...$$ upto 40 terms. If $$(\tan\beta)^{\frac{\alpha}{1020}}$$ is a root of the equation $$x^2 + x - 2 = 0$$, $$\beta \in \left(0, \frac{\pi}{2}\right)$$, then $$\sin^2\beta + 3\cos^2\beta$$ is equal to :

$$\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots \text{ upto 40 terms}$$

$$\alpha = (3 + 4) + (8 + 9) + (13 + 14) + \dots \text{ upto 20 pairs}$$

$$\alpha = 7 + 17 + 27 + \dots \text{ upto 20 terms}$$

This is an AP with first term $$a = 7$$, common difference $$d = 10$$, and number of terms $$n = 20$$.

$$\alpha = \frac{n}{2} [2a + (n - 1)d]$$

$$\alpha = \frac{20}{2} [2(7) + (20 - 1)10]$$

$$\alpha = 10 [14 + 190] = 10 \times 204 = 2040$$

$$x^2 + x - 2 = 0$$

$$(x + 2)(x - 1) = 0 \implies x = 1 \text{ or } x = -2$$

$$x = (\tan \beta)^{\frac{\alpha}{1020}} = 1$$

$$(\tan \beta)^{\frac{2040}{1020}} = 1$$

$$(\tan \beta)^2 = 1 \implies \tan \beta = 1 \quad \left[\text{since } \beta \in \left(0, \frac{\pi}{2}\right)\right]$$

$$\beta = \frac{\pi}{4}$$

$$\sin^2 \beta + 3 \cos^2 \beta = \sin^2\left(\frac{\pi}{4}\right) + 3 \cos^2\left(\frac{\pi}{4}\right)$$

$$\sin^2 \beta + 3 \cos^2 \beta = \left(\frac{1}{\sqrt{2}}\right)^2 + 3\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$$