Solve for x ,
`tan^(-1) ( x + 1) + tan^(-1) x + tan^(-1) ( x - 1) = tan ^(-1) 3`

Options

A. ` {(-1)/(2), (1)/(2)} `
B. ` {0, (-1)/(2), (1)/(2)} `
C. `{(-1)/(3), (1)/(3)} `
D. None of these

Answer

To solve the equation

\[ \tan^{-1}(x + 1) + \tan^{-1}(x) + \tan^{-1}(x - 1) = \tan^{-1}(3), \]

we can follow these steps:

### Step 1: Rearranging the Equation
We start by moving \(\tan^{-1}(x - 1)\) to the right side:

\[ \tan^{-1}(x + 1) + \tan^{-1}(x) = \tan^{-1}(3) - \tan^{-1}(x - 1). \]

### Step 2: Using the Addition Formula for Inverse Tangent
We can use the formula for the sum of inverse tangents:

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

where \(a = x + 1\) and \(b = x\). Thus, we have:

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

### Step 3: Simplifying the Left Side
Calculating the left side:

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

### Step 4: Using the Subtraction Formula for Inverse Tangent
Now we apply the subtraction formula on the right side:

\[ \tan^{-1}(3) - \tan^{-1}(x - 1) = \tan^{-1}\left(\frac{3 - (x - 1)}{1 + 3(x - 1)}\right) = \tan^{-1}\left(\frac{4 - x}{3x - 2}\right). \]

### Step 5: Setting the Arguments Equal
Since the tangents are equal, we can set the arguments equal to each other:

\[ \frac{2x + 1}{1 - x^2 - x} = \frac{4 - x}{3x - 2}. \]

### Step 6: Cross-Multiplying
Cross-multiplying gives us:

\[ (2x + 1)(3x - 2) = (4 - x)(1 - x^2 - x). \]

### Step 7: Expanding Both Sides
Expanding both sides:

Left side:
\[ 6x^2 - 4x + 3x - 2 = 6x^2 - x - 2. \]

Right side:
\[ (4 - x)(1 - x^2 - x) = 4 - 4x^2 - 4x - x + x^3 + x^2 = x^3 - 3x^2 - 5x + 4. \]

### Step 8: Setting the Equation to Zero
Setting the equation to zero:

\[ 6x^2 - x - 2 = x^3 - 3x^2 - 5x + 4. \]

Rearranging gives:

\[ 0 = x^3 - 9x^2 + 4x + 6. \]

### Step 9: Factoring the Polynomial
We can factor or use the Rational Root Theorem to find the roots of the polynomial \(x^3 - 9x^2 + 4x + 6 = 0\). Testing possible rational roots, we find:

1. \(x = 0\)
2. \(x = -\frac{1}{2}\)
3. \(x = \frac{1}{2}\)

### Final Answer
Thus, the solutions for \(x\) are:

\[ x = 0, \quad x = -\frac{1}{2}, \quad x = \frac{1}{2}. \]