Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f:(-1,1)\to B$, be a function defined by $f(x)={\tan }^{-1}(\frac{2x}{1-x^2})$, then $f$ is both one - one and onto when $B$ is the interval

Imagine you are standing at the edge of a complex mathematical landscape. You see a function, $f(x)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$, and it looks intimidating.

It is a composite function, a tangle of algebra and trigonometry. But in the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to find the codomain $B$ that makes this function bijective—both one-one and onto.

Unmasking the Identity

The first step in any great problem is to recognize the pattern. Look closely at the expression inside the inverse tangent:

Does it ring a bell? It is the ghost of the double-angle identity for tangent:

This is our key. We make the substitution $x=\tan \theta$.

But we must be careful! The problem explicitly states that $x\in (-1,1)$. This is not just a constraint; it is a boundary condition.

If $x$ is between $-1$ and $1$, then $\theta$ must be between $-\frac{\pi }{4}$ and $\frac{\pi }{4}$. This restriction is the guardian of our logic.

The Trap

Now, we substitute. The function becomes $f(x)={\tan }^{-1}(\tan 2\theta )$.

Here is where many students stumble. They blindly cancel the ${\tan }^{-1}$ and $\tan$ to get $2\theta$. But in mathematics, we must always check the principal value branch.

Since $\theta \in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)$, it follows that $2\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. This interval is exactly the principal domain of the inverse tangent function.

Because our angle $2\theta$ sits comfortably within this safe zone, we can indeed simplify the expression to $f(x)=2\theta$. Substituting back, we find the true, simplified soul of our function:

The One-to-One Proof

Now that we have $f(x)=2{\tan }^{-1}x$, the path forward is clear. To prove a function is one-one, we must show it is strictly monotonic.

Let us look at the derivative:

Notice the denominator: $1+x^2$. Since $x^2$ is always non-negative, $1+x^2$ is always positive.

Thus, $f^{\prime }(x)>0$ for all $x$ in the domain. A strictly positive derivative means our function is strictly increasing. It never looks back, never repeats a value; it is, undeniably, one-to-one.

Range and Onto

We have conquered the one-to-one property. Now, for the final hurdle: the onto property.

A function is onto if and only if its codomain equals its range. We need to find the range of $f(x)=2{\tan }^{-1}x$ over the domain $x\in (-1,1)$.

Since ${\tan }^{-1}x$ is a strictly increasing function, we can apply it to the inequality $-1<x<1$ without flipping the signs. We get:

This simplifies to $-\frac{\pi }{4}<{\tan }^{-1}x<\frac{\pi }{4}$. Multiplying by $2$, we arrive at the range:

This is our interval $B$. The function is onto when the codomain $B$ is exactly this range: $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.