Animated Solution for Mathematics - Inverse Trigonometric Functions: Let ${\tan }^{-1}y={\tan }^{-1}x+{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$, where $|x|<\frac{1}{\sqrt{3}}$. Then a value of $y$ is :

Welcome, my dear student. Today, we are going to peel back the layers of a problem that, at first glance, might seem like a simple algebraic substitution, but is actually a beautiful exercise in understanding the domain of inverse trigonometric functions.

We are given the equation ${\tan }^{-1}y={\tan }^{-1}x+{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ with the crucial constraint $|x|<\frac{1}{\sqrt{3}}$.

Many students see this constraint and ignore it, treating it as mere background noise. But in the world of JEE Advanced, the constraint is the soul of the problem. It tells us exactly which branch of the inverse tangent function we are operating in.

Let us look at the right-hand side of our equation. The term ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ should immediately ring a bell. It is the classic identity for $2{\tan }^{-1}x$.

However, we must pause. Is it always $2{\tan }^{-1}x$? Not quite. This identity is valid only when $|x|<1$.

Since our problem explicitly states $|x|<\frac{1}{\sqrt{3}}$, and we know that $\frac{1}{\sqrt{3}}\approx 0.577$, which is clearly less than $1$, we are in the safe zone. We can proceed with the substitution without any fear of branch shifts.

Our equation now transforms into something much more manageable:

Now, look at how the complexity melts away. We are simply adding like terms. Just as $a+2a=3a$, our equation becomes:

We have successfully reduced a daunting inverse trigonometric expression into a compact, elegant form. But we are not done yet. We need to find $y$.

To do this, we must express $3{\tan }^{-1}x$ as a single inverse tangent function. This brings us to the triple angle identity:

Again, we check our constraint. This identity is valid for $|x|<\frac{1}{\sqrt{3}}$. It is a perfect match! The problem designer has carefully chosen this range to ensure we do not need to worry about adding or subtracting $\pi$.

With the identity in hand, we substitute it back:

Since the inverse tangent function is one-to-one within its principal domain, we can equate the arguments directly. Thus, we arrive at our final result:

This matches the third option provided in the problem. The lesson here is profound: never rush into calculations. Take a moment to observe the constraints, recognize the standard identities, and let the structure of the mathematics guide you to the solution.