Solve for x, `cos^(-1)((x^(2)-1)/(x^(2)+1))+tan^(-1)((-2x)/(1-x^(2)))=(2pi)/(3)`

Answer

To solve the equation

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

we will follow these steps:

### Step 1: Substitute \(x\) with \(\tan(\theta)\)
Let \(x = \tan(\theta)\). Then, we can rewrite the expressions:

\[ \frac{x^2 - 1}{x^2 + 1} = \frac{\tan^2(\theta) - 1}{\tan^2(\theta) + 1} = \frac{\sec^2(\theta) - 2}{\sec^2(\theta)} = \cos(2\theta) \]

and

\[ \frac{-2x}{1 - x^2} = \frac{-2\tan(\theta)}{1 - \tan^2(\theta)} = -\tan(2\theta). \]

### Step 2: Rewrite the equation
Substituting these into the original equation gives:

\[ \cos^{-1}(\cos(2\theta)) + \tan^{-1}(-\tan(2\theta)) = \frac{2\pi}{3}. \]

### Step 3: Simplify using properties of inverse trigonometric functions
Using the properties of inverse trigonometric functions, we have:

\[ \cos^{-1}(\cos(2\theta)) = 2\theta \quad \text{(for } 0 \leq 2\theta \leq \pi\text{)} \]
and
\[ \tan^{-1}(-\tan(2\theta)) = -\tan^{-1}(\tan(2\theta)) = -2\theta. \]

Thus, the equation simplifies to:

\[ 2\theta - 2\theta = \frac{2\pi}{3}. \]

### Step 4: Solve for \(\theta\)
This simplifies to:

\[ \pi - 4\theta = \frac{2\pi}{3}. \]

Rearranging gives:

\[ 4\theta = \pi - \frac{2\pi}{3} = \frac{\pi}{3}. \]

Dividing both sides by 4:

\[ \theta = \frac{\pi}{12}. \]

### Step 5: Find \(x\)
Since we set \(x = \tan(\theta)\), we have:

\[ x = \tan\left(\frac{\pi}{12}\right). \]

### Final Answer
Thus, the solution for \(x\) is:

\[ x = \tan\left(\frac{\pi}{12}\right). \]

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