Solve the following equations :
`tan^(-1) ((x-2)/(x-4)) +tan^(-1) ((x+2)/(x+4)) = pi/ 4 `

Answer

To solve the equation

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

we can use the formula for the sum of inverse tangents:

\[ \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A + B}{1 - AB} \right), \]

provided that \(AB < 1\).

### Step 1: Identify A and B
Let:
\[ A = \frac{x-2}{x-4}, \quad B = \frac{x+2}{x+4}. \]

### Step 2: Apply the formula
Using the formula, we have:

\[ \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A + B}{1 - AB} \right) = \frac{\pi}{4}. \]

This implies:

\[ \frac{A + B}{1 - AB} = 1. \]

### Step 3: Set up the equation
This leads to the equation:

\[ A + B = 1 - AB. \]

### Step 4: Calculate A + B
Calculate \(A + B\):

\[ A + B = \frac{x-2}{x-4} + \frac{x+2}{x+4}. \]

Finding a common denominator:

\[ A + B = \frac{(x-2)(x+4) + (x+2)(x-4)}{(x-4)(x+4)}. \]

Expanding the numerator:

\[ = \frac{(x^2 + 4x - 2x - 8) + (x^2 - 4x + 2x - 8)}{x^2 - 16} = \frac{2x^2 - 16}{x^2 - 16}. \]

### Step 5: Calculate AB
Now calculate \(AB\):

\[ AB = \left( \frac{x-2}{x-4} \right) \left( \frac{x+2}{x+4} \right) = \frac{(x-2)(x+2)}{(x-4)(x+4)} = \frac{x^2 - 4}{x^2 - 16}. \]

### Step 6: Substitute A + B and AB into the equation
Substituting \(A + B\) and \(AB\) into the equation \(A + B = 1 - AB\):

\[ \frac{2x^2 - 16}{x^2 - 16} = 1 - \frac{x^2 - 4}{x^2 - 16}. \]

### Step 7: Simplify the equation
Cross-multiplying gives:

\[ 2x^2 - 16 = (x^2 - 16) - (x^2 - 4) = -12. \]

### Step 8: Solve for x
Now we have:

\[ 2x^2 - 16 = -12 \implies 2x^2 = 4 \implies x^2 = 2. \]

Taking the square root:

\[ x = \pm \sqrt{2}. \]

### Final Solution
Thus, the solutions to the equation are:

\[ x = \sqrt{2} \quad \text{and} \quad x = -\sqrt{2}. \]