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

Unraveling Trigonometric Series

Welcome, future engineers! Today, we are going to embark on a journey through a problem that, at first glance, might seem like a chaotic mess of infinite series.

When you see a summation symbol ${\sum }_{n=0}^{\infty }$ in a JEE Advanced paper, it is natural to feel a momentary spike of anxiety. But I want you to take a deep breath. In the world of competitive mathematics, an infinite series is often just a finite, elegant structure hiding in plain sight.

Our goal today is to peel back the layers of these trigonometric expressions and find the hidden bond between $x$ and $y$.

Decoding the Series for $x$

Let us look at our first expression: $x={\sum }_{n=0}^{\infty }(-1{)}^n{\tan }^{2n}\theta$. When we expand this, we see $1-{\tan }^2\theta +{\tan }^4\theta -{\tan }^6\theta +\dots$.

Every term is generated by multiplying the previous one by $-{\tan }^2\theta$. This is the classic signature of an infinite Geometric Progression (G.P.).

In any G.P., the sum to infinity is given by $S_{\infty }=\frac{a}{1-r}$, provided the common ratio $|r|<1$. Here, our first term $a=1$ and our common ratio $r=-{\tan }^2\theta$.

Because we are given the constraint $0<\theta <\frac{\pi }{4}$, we know that $\tan \theta$ is between $0$ and $1$, which means ${\tan }^2\theta$ is also between $0$ and $1$. The convergence condition is satisfied!

Substituting these into our formula, we get:

Now, recall the fundamental trigonometric identity $1+{\tan }^2\theta ={\sec }^2\theta$. Substituting this in, we get $x=\frac{1}{{\sec }^2\theta }$.

Since $\frac{1}{\sec \theta }=\cos \theta$, we have arrived at our first beautiful result: $x={\cos }^2\theta$. Keep this safe; it is the key to our puzzle.

Decoding the Series for $y$

Now, let us turn our attention to $y={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta$. Expanding this, we get $1+{\cos }^2\theta +{\cos }^4\theta +{\cos }^6\theta +\dots$.

Again, this is an infinite G.P., but this time with a first term $a=1$ and a common ratio $r={\cos }^2\theta$. Since $0<\theta <\frac{\pi }{4}$, we know that $0<{\cos }^2\theta <1$, so the series converges perfectly.

Applying the sum formula again, we get:

This denominator should immediately ring a bell. We know that ${\sin }^2\theta +{\cos }^2\theta =1$, which means $1-{\cos }^2\theta ={\sin }^2\theta$.

Therefore, $y=\frac{1}{{\sin }^2\theta }$. If we rearrange this to isolate ${\sin }^2\theta$, we get ${\sin }^2\theta =\frac{1}{y}$.

The Synthesis

We have now reduced two intimidating infinite series into two simple trigonometric expressions: $x={\cos }^2\theta$ and ${\sin }^2\theta =\frac{1}{y}$. How do we connect them?

We use the most powerful tool in our trigonometric arsenal: the Pythagorean identity, ${\sin }^2\theta +{\cos }^2\theta =1$.

Substituting our expressions into this identity, we get:

We are almost at the finish line! To match the format of the options provided, we need to clear the fraction. Multiply the entire equation by $y$:

Now, rearrange the terms to group the $y$ variables: $y-xy=1$. Finally, factor out $y$ to get the final relationship:

Look at what we have achieved. We started with infinite sums that seemed to stretch on forever, and through the power of G.P. formulas and trigonometric identities, we distilled them into a simple, elegant algebraic relationship.

This is the essence of JEE Advanced mathematics: taking complexity and finding the underlying simplicity. Never let the notation intimidate you; look for the structure, trust your identities, and the answer will reveal itself.