Evaluate Trigonometric Expression Using Series Sum Root

To find the value of the given expression, we first evaluate the sum of the series $\alpha$.

The given series is: $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots \text{ upto } 40 \text{ terms}$

We can group the terms in pairs: $\alpha = (3 + 4) + (8 + 9) + (13 + 14) + \dots \text{ upto } 20 \text{ pairs}$ $\alpha = 7 + 17 + 27 + \dots \text{ upto } 20 \text{ terms}$

This is an Arithmetic Progression (AP) with:

• First term, $a = 7$
• Common difference, $d = 10$
• Number of terms, $N = 20$

Using the formula for the sum of an AP, $S_N = \frac{N}{2} [2a + (N-1)d]$: $\alpha = \frac{20}{2} \left[2(7) + (20 - 1)10\right]$ $\alpha = 10 \left[14 + 190\right] = 10 \times 204 = 2040$

Next, we evaluate the exponent: $\frac{\alpha}{1020} = \frac{2040}{1020} = 2$

So, the root given in the problem is $(\tan\beta)^2$.

It is given that $(\tan\beta)^2$ is a root of the quadratic equation: $x^2 + x - 2 = 0$

Solving the quadratic equation: $(x + 2)(x - 1) = 0 \implies x = 1 \quad \text{or} \quad x = -2$

Since $\beta \in \left(0, \frac{\pi}{2}\right)$, we have $\tan\beta > 0$, which implies $(\tan\beta)^2 > 0$. Thus, we reject the negative root $x = -2$: $(\tan\beta)^2 = 1$

Since $\beta \in \left(0, \frac{\pi}{2}\right)$, taking the positive square root yields: $\tan\beta = 1 \implies \beta = \frac{\pi}{4}$

Now, we evaluate the expression $\sin^2\beta + 3\cos^2\beta$: $\sin^2\left(\frac{\pi}{4}\right) + 3\cos^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 + 3\left(\frac{1}{\sqrt{2}}\right)^2$ $= \frac{1}{2} + 3\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{3}{2} = 2$

Thus, the value of $\sin^2\beta + 3\cos^2\beta$ is equal to $2$.