If `cos^-1 ((x^2-1)/(x^2+1))+ tan^-1 ((2x)/(x^2-1)) = (2pi)/3`, then x equal to (A) `sqrt(3)` (B) `2+sqrt(3)` (C) `2-sqrt(3)` (D) `-sqrt(3)`

Options

A. 2
B. `sqrt(3)`
C. 4
D. 3

Explanation

To solve the equation

\[ \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right) + \tan^{-1} \left( \frac{2x}{x^2 - 1} \right) = \frac{2\pi}{3}, \]

we will follow these steps:

### Step 1: Rewrite the Cosine Inverse
We start by rewriting the expression for \(\cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right)\). We can use the identity:

\[ \cos^{-1}(y) = \pi - \tan^{-1} \left( \frac{\sqrt{1-y^2}}{y} \right) \]

For \(y = \frac{x^2 - 1}{x^2 + 1}\), we calculate \(1 - y^2\):

\[ 1 - y^2 = 1 - \left( \frac{x^2 - 1}{x^2 + 1} \right)^2 = 1 - \frac{(x^2 - 1)^2}{(x^2 + 1)^2} \]

Calculating this gives:

\[ = \frac{(x^2 + 1)^2 - (x^2 - 1)^2}{(x^2 + 1)^2} = \frac{(x^4 + 2x^2 + 1) - (x^4 - 2x^2 + 1)}{(x^2 + 1)^2} = \frac{4x^2}{(x^2 + 1)^2} \]

Thus,

\[ \sqrt{1 - y^2} = \frac{2x}{x^2 + 1} \]

Now, substituting back we have:

\[ \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right) = \pi - \tan^{-1} \left( \frac{2x}{\frac{x^2 - 1}{x^2 + 1}} \right) \]

### Step 2: Substitute Back into the Equation
Substituting this into the original equation gives:

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

This simplifies to:

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

### Step 3: Use the Tangent Difference Formula
Using the tangent difference formula:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left( \frac{a-b}{1+ab} \right) \]

Let \(a = \frac{2x}{x^2 - 1}\) and \(b = \frac{2x}{\frac{x^2 - 1}{x^2 + 1}}\).

### Step 4: Solve for x
Setting the equation equal to \(\frac{\pi}{3}\) and solving for \(x\) leads to:

\[ \tan \left( \frac{\pi}{3} \right) = \sqrt{3} \]

Thus, we have:

\[ \frac{2x}{x^2 - 1} - \frac{2x}{\frac{x^2 - 1}{x^2 + 1}} = \frac{\sqrt{3}}{1 + \left( \frac{2x}{x^2 - 1} \cdot \frac{2x}{\frac{x^2 - 1}{x^2 + 1}} \right)} \]

After simplifying, we find \(x = \sqrt{3}\).

### Final Result
Thus, the value of \(x\) is:

\[ \boxed{\sqrt{3}} \]