Animated Solution for Mathematics - Trigonometry: If $\exp \{({\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \infty )\ln 2\}$ satisfies the equation $x^2-9x+8=0$, find the value of $\frac{\cos x}{\cos x+\sin x},0<x<\pi /2$.

Welcome, fellow traveler of the JEE journey. Today, we stand before a problem that, at first glance, appears to be a chaotic mess of exponentials, infinite series, and quadratic equations. But I want you to take a deep breath. In mathematics, as in physics, complexity is often just a mask for underlying symmetry. Let us peel back that mask together.

Our journey begins with the exponent: $S={\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+\dots \infty$. Do you see it? This is not just a random collection of terms; it is the heartbeat of an infinite geometric series.

In a geometric series, each term is generated by multiplying the previous one by a constant ratio, $r$. Here, our first term is $a={\sin }^{2}x$, and our common ratio is also $r={\sin }^{2}x$. Because our angle $x$ is constrained within the open interval $(0,\frac{\pi }{2})$, we know that $0<{\sin }^{2}x<1$. This is the green light we need—the series is guaranteed to converge!

Now that we know the series converges, we invoke the classic sum formula for an infinite geometric progression:

Substituting our values, we get $S=\frac{{\sin }^{2}x}{1-{\sin }^{2}x}$. This is where the magic happens. We reach into our trigonometric toolkit and pull out the most fundamental identity of all: ${\sin }^{2}x+{\cos }^{2}x=1$.

This allows us to rewrite the denominator as ${\cos }^{2}x$. Suddenly, the expression simplifies beautifully:

The infinite series has collapsed into a single, elegant term: ${\tan }^{2}x$.

With the exponent tamed, our original expression becomes $\exp \{{\tan }^{2}x\ln 2\}$. Using the logarithmic power rule, where $a\ln b=\ln (b^a)$, we rewrite this as $e^{\ln (2^{{\tan }^{2}x})}$.

Since the exponential function and the natural logarithm are inverse functions, they cancel each other out, leaving us with the clean, manageable expression: $2^{{\tan }^{2}x}$.

Now, the problem tells us this expression satisfies the quadratic equation $y^2-9y+8=0$, where $y=2^{{\tan }^{2}x}$. Factoring this quadratic is straightforward: $(y-1)(y-8)=0$. This gives us two potential paths: $y=1$ or $y=8$. We must test both.

If $2^{{\tan }^{2}x}=1$, then ${\tan }^{2}x=0$, which implies $x=0$. But look back at our constraints! The problem explicitly defines $0<x<\frac{\pi }{2}$. Our first root is an imposter; we must reject it.

That leaves us with the second path: $2^{{\tan }^{2}x}=8$. Since $8=2^3$, we equate the exponents: ${\tan }^{2}x=3$. Taking the square root (and keeping the positive value because $x$ is in the first quadrant), we find $\tan x=\sqrt{3}$. This corresponds to the standard angle $x=\frac{\pi }{3}$, or $60^{\circ }$.

We have arrived at the final stage. We need to evaluate the expression $\frac{\cos x}{\cos x+\sin x}$ at $x=\frac{\pi }{3}$. We know that $\cos (\frac{\pi }{3})=\frac{1}{2}$ and $\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$.

Substituting these values, we get:

To finish with elegance, we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{3}-1)$. The result is:

Look at what you have achieved. You took a terrifying exponential expression, tamed an infinite series, solved a quadratic, and navigated trigonometric constraints to reach a precise, beautiful answer. This is the essence of JEE Advanced—not just calculation, but the art of simplification.