Animated Solution for Mathematics - Trigonometry: If the range of $f(\theta )=\frac{{\sin }^4\theta +3{\cos }^2\theta }{{\sin }^4\theta +{\cos }^2\theta },\theta \in \mathbb{R}$ is $[\alpha ,\beta ]$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{\alpha }{\beta }$, is equal to ________

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of powers and trigonometric ratios. You see an expression like $f(\theta )=\frac{{\sin }^4\theta +3{\cos }^2\theta }{{\sin }^4\theta +{\cos }^2\theta }$ and your instinct might be to panic.

But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for a hidden, elegant simplicity. Our job is to peel back that mask.

Look closely at the numerator: ${\sin }^4\theta +3{\cos }^2\theta$. Now look at the denominator: ${\sin }^4\theta +{\cos }^2\theta$. Do you see the symmetry?

The numerator is almost identical to the denominator. This is our entry point. We don't need complex identities; we need simple arithmetic. Let's perform some 'algebraic surgery' by splitting the numerator:

By separating the fraction, we get:

Suddenly, the problem feels lighter, doesn't it? We have isolated the constant $1$, and now we only need to focus on the behavior of the remaining fraction.

Trigonometry is beautiful, but sometimes it gets in the way of clear algebraic thinking. Let's introduce a dummy variable to clear the fog. Let $t={\cos }^2\theta$.

Since $\cos \theta$ oscillates between $-1$ and $1$, its square, $t$, is strictly confined to the interval $[0,1]$.

Now, let's transform the denominator. We know that ${\sin }^2\theta =1-{\cos }^2\theta =1-t$. Therefore, ${\sin }^4\theta =(1-t{)}^2$. Substituting this into our denominator:

Our function is now a clean, rational algebraic function of $t$:

We need the range of $g(t)$. Let's check the boundaries first. At $t=0$, $g(0)=1+0=1$. This is our minimum.

Now, what happens when $t>0$? To analyze the fraction $\frac{2t}{t^2-t+1}$, let's divide both the numerator and the denominator by $t$:

This is where the magic happens. We have the expression $t+\frac{1}{t}$. As we discussed in our FAQs, for any $t>0$, the sum of a number and its reciprocal is at least $2$.

Thus, the minimum value of $t+\frac{1}{t}$ is $2$. Consequently, the minimum value of the denominator $t+\frac{1}{t}-1$ is $2-1=1$.

Since the denominator's minimum is $1$, the maximum value of the fraction $\frac{2}{t+\frac{1}{t}-1}$ is $\frac{2}{1}=2$. Adding the $1$ we had outside, the maximum value of our function $g(t)$ is $1+2=3$.

We have successfully bounded our function: the range is $[1,3]$. Therefore, $\alpha =1$ and $\beta =3$.

We aren't done yet! The problem asks for the sum of an infinite geometric progression (G.P.) where the first term $a=64$ and the common ratio $r=\frac{\alpha }{\beta }=\frac{1}{3}$.

The formula for the sum of an infinite G.P. is a classic:

Plugging in our values:

And there it is. 96. You started with a daunting trigonometric expression, performed a substitution, analyzed a rational function, and finished with a clean, elegant result.

This is the essence of JEE Advanced physics and mathematics—it is not about memorizing formulas; it is about seeing the structure, simplifying the chaos, and trusting your tools. Keep practicing this mindset, and you will be unstoppable.