JEE Mathematics 2023 Question with Solution

Given:

$$\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}$$

with $-1 < x < 1$ and $x \neq 0$.

Find: $\alpha$ if the sum of all solutions is $\alpha - \frac{4}{\sqrt{3}}$.

Using the inverse trigonometric identity carefully by cases:

For Case I: $x > 0$,

$$\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$$

So,

$$2\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}$$

Hence,

$$\tan^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{6}$$

Therefore,

$$\frac{2x}{1-x^2}=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}$$

which gives

$$2\sqrt{3}x=1-x^2$$ $$x^2+2\sqrt{3}x-1=0$$

Solving,

$$x=-\sqrt{3}\pm 2$$

From the interval $-1<x<1$ and the case $x>0$, the valid solution is

$$x=2-\sqrt{3}$$

For Case II: $x < 0$,

$$\cot^{-1}\left(\frac{1-x^2}{2x}\right)=\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\pi$$

So,

$$2\tan^{-1}\left(\frac{2x}{1-x^2}\right)+\pi=\frac{\pi}{3}$$

This yields the negative-branch solution reported in the extracted working as

$$x=-\frac{1}{\sqrt{3}}$$

Hence the sum of all solutions is

$$\left(2-\sqrt{3}\right)-\frac{1}{\sqrt{3}}=2-\frac{4}{\sqrt{3}}$$

Comparing with $\alpha-\frac{4}{\sqrt{3}}$, we get

$$\alpha=2$$

Therefore, the required value of $\alpha$ is $2$.

Given:

$$\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}$$

Find: $\alpha$.

Notice that

$$\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$$

So the expression

$$\frac{2x}{1-x^2}$$

has the standard form with $x=\tan\theta$.

The extracted solution handles the inverse-function ranges separately for $x>0$ and $x<0$. From those range adjustments, the valid solutions in the interval $$-1

$$$$