Animated Solution for Mathematics - Trigonometry: If $x={\sum }_{n=0}^{\infty }(-1{)}^n{\tan }^{2n}\theta$ and $y={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta$, where $0<\theta <\frac{\pi }{4}$, then:

Imagine you are standing on the edge of an infinite abyss, looking at a sequence of numbers that never ends. To a novice, this looks like chaos. To a JEE aspirant, this is a playground.

Today, we are going to tame two infinite series, $x$ and $y$, and uncover the hidden geometric harmony connecting them. We are given:

We operate under the crucial constraint $0<\theta <\frac{\pi }{4}$.

Let us first dissect $y$. When we expand the summation $y={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta$, we get $y=1+{\cos }^2\theta +{\cos }^4\theta +\dots$.

This is a classic infinite Geometric Progression (GP) where the first term $a=1$ and the common ratio $r={\cos }^2\theta$. Because our angle $\theta$ is restricted between $0$ and $\frac{\pi }{4}$, we know that $0<{\cos }^2\theta <1$.

This is the green light we need! The sum of an infinite GP is given by $S_{\infty }=\frac{a}{1-r}$. Substituting our values:

Using the fundamental identity ${\sin }^2\theta +{\cos }^2\theta =1$, we can rewrite the denominator as ${\sin }^2\theta$. Thus, $y=\frac{1}{{\sin }^2\theta }$, or more conveniently:

Now, let us turn our attention to $x={\sum }_{n=0}^{\infty }(-1{)}^n{\tan }^{2n}\theta$. Expanding this, we see $x=1-{\tan }^2\theta +{\tan }^4\theta -\dots$.

Notice the alternating signs? This is another infinite GP, but with a twist. Here, the first term $a=1$, and the common ratio $r=-{\tan }^2\theta$.

Again, because $\theta <\frac{\pi }{4}$, we have ${\tan }^2\theta <1$, so the series converges. Applying the sum formula $S_{\infty }=\frac{a}{1-r}$:

Recalling the identity $1+{\tan }^2\theta ={\sec }^2\theta$, we find $x=\frac{1}{{\sec }^2\theta }$, which simplifies beautifully to:

We have reduced two intimidating infinite series into two elegant trigonometric expressions: ${\sin }^2\theta =\frac{1}{y}$ and $x={\cos }^2\theta$. Now, we invoke the most powerful bridge in trigonometry: ${\sin }^2\theta +{\cos }^2\theta =1$.

Substituting our expressions, we get:

With a simple algebraic shuffle—subtracting $x$ from both sides and rearranging—we arrive at the final relationship:

Look at that! The infinite complexity has vanished, leaving behind a simple, elegant relationship. This is the essence of JEE mathematics: identifying the underlying structure, applying the right tools, and watching the complexity collapse into simplicity.