Animated Solution for Mathematics - Inverse Trigonometric Functions: If the sum of all the solutions of ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\pi }{3}$, $-1<x<1,x\ne 0$, is $\alpha -\frac{4}{\sqrt{3}}$, then $\alpha$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are unraveling a geometric mystery.

We are presented with the equation:

This is subject to the strict constraint that $-1<x<1$ and $x eq 0$. This is a classic problem that tests your understanding of the subtle, often hidden, behavior of inverse trigonometric functions.

Before we touch the algebra, we must respect the domain. The number line is our battlefield, and the point $x=0$ is our dividing line.

The function ${\cot }^{-1}(y)$ behaves differently depending on whether its argument $y$ is positive or negative. Our argument is $y=\frac{1-x^2}{2x}$.

Since $|x|<1$, the numerator $1-x^2$ is always positive. Thus, the sign of $y$ is entirely determined by the sign of $x$. We must split our analysis into two distinct cases: $x\in (0,1)$ and $x\in (-1,0)$.

The Positive Territory

Imagine we are in the region where $x$ is positive ($0<x<1$). In this realm, our argument $y=\frac{1-x^2}{2x}$ is positive.

The identity for inverse cotangent is:

Substituting our $y$, we get ${\cot }^{-1}\left(\frac{1-x^2}{2x}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$. Our original equation transforms into:

This simplifies to $2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$. Using the standard identity ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=2{\tan }^{-1}x$, we obtain:

Taking the tangent of both sides, we find $x=\tan \left(\frac{\pi }{12}\right)=2-\sqrt{3}$. This value is approximately $0.268$, which sits comfortably within our interval $(0,1)$.

The Negative Territory

Now, let us step into the negative side, where $-1<x<0$. Here, the argument $y=\frac{1-x^2}{2x}$ is negative.

This is where the JEE trap lies! The identity shifts to:

Our equation becomes:

Combining the terms, we have $2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+\pi =\frac{\pi }{3}$. This leads to:

Dividing by 4, we get ${\tan }^{-1}x=-\frac{\pi }{6}$. Thus, $x=\tan \left(-\frac{\pi }{6}\right)=-\frac{1}{\sqrt{3}}$. This value is approximately $-0.577$, which perfectly satisfies our condition $-1<x<0$.

We have two valid solutions: $x_1=2-\sqrt{3}$ and $x_2=-\frac{1}{\sqrt{3}}$. The problem asks for the sum of all solutions.

Let us calculate:

Comparing this to the form $\alpha -\frac{4}{\sqrt{3}}$, we see that $\alpha =2$. We have navigated the domain, avoided the identity traps, and arrived at the truth.