Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $S=\{x\in \mathbb{R}:0<x<1\text{ and }2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\}$. If $n(S)$ denotes the number of elements in $S$ then :

Imagine you are standing before a complex equation. It looks intimidating, doesn't it? We have $2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ with the constraint $0<x<1$.

Many students see this and immediately reach for derivatives or complex algebraic expansions. But stop—take a breath. In the world of JEE Advanced, the most complex-looking problems are often hiding a beautiful, elegant symmetry.

Whenever you see expressions like $\frac{1-x}{1+x}$ or $\frac{1-x^2}{1+x^2}$, your mathematical intuition should scream, "Trigonometric Substitution!" These are not random fractions; they are the ghosts of the tangent addition and double-angle formulas.

Let us set $x=\tan \theta$. Since our domain is $0<x<1$, our angle $\theta$ must live in the interval $(0,\frac{\pi }{4})$. This is our playground, and we must stay within these bounds to ensure our identities remain valid.

Let us tackle the left side: $2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)$. Substituting $x=\tan \theta$, we get:

Does that inner term look familiar? It is the expansion of $\tan (\frac{\pi }{4}-\theta )$. So, the expression becomes $2{\tan }^{-1}(\tan (\frac{\pi }{4}-\theta ))$.

Because $\theta$ is in $(0,\frac{\pi }{4})$, the angle $(\frac{\pi }{4}-\theta )$ is also in $(0,\frac{\pi }{4})$. This is perfect! It falls right into the principal range of the inverse tangent function, allowing us to simplify it directly to:

Now, let us turn our attention to the right side: ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$. Again, substituting $x=\tan \theta$, we obtain:

This is the classic double-angle identity for cosine: $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$. Thus, we have ${\cos }^{-1}(\cos 2\theta )$.

Since $\theta \in (0,\frac{\pi }{4})$, it follows that $2\theta \in (0,\frac{\pi }{2})$. This range is well within the principal domain of the cosine inverse function, so ${\cos }^{-1}(\cos 2\theta )$ simplifies cleanly to $2\theta$.

We have arrived at the moment of truth. Equating our simplified sides, we get:

This is a simple linear equation! Adding $2\theta$ to both sides, we find $4\theta =\frac{\pi }{2}$, which means $\theta =\frac{\pi }{8}$.

Now, we must return to our original variable $x$. Since $x=\tan \theta$, we have $x=\tan (\frac{\pi }{8})$. Using the half-angle formula, we know that:

Since $\sqrt{2}\approx 1.414$, our value for $x$ is approximately $0.414$.

Look at the options provided. We found exactly one value for $x$, so $n(S)=1$. Furthermore, since $0.414<0.5$, we can confidently say that the element in $S$ is less than $\frac{1}{2}$.

See how the complexity melted away? We didn't need brute force; we needed the right perspective. By transforming the algebraic mess into a trigonometric dance, we found the solution with grace.