If `tan^(-1)((x-2)/(x-4))+tan^(-1)((x+2)/(x+4))=(pi)/(4)`, find the value of 'x'.

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 two 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

\[ a = \frac{x-2}{x-4} \quad \text{and} \quad b = \frac{x+2}{x+4}. \]

### Step 2: Calculate \(a + b\)

Now, we calculate \(a + b\):

\[ a + b = \frac{x-2}{x-4} + \frac{x+2}{x+4}. \]

To add these fractions, we need a common denominator:

\[ a + b = \frac{(x-2)(x+4) + (x+2)(x-4)}{(x-4)(x+4)}. \]

### Step 3: Expand the numerator

Expanding the numerator:

\[ (x-2)(x+4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8, \]
\[ (x+2)(x-4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8. \]

Now, combine these:

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

### Step 4: Calculate \(ab\)

Next, we 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 5: Substitute into the inverse tangent formula

Now, substituting \(a + b\) and \(ab\) into the inverse tangent formula:

\[ \tan^{-1}\left(\frac{2x^2 - 16}{(x-4)(x+4)(1 - \frac{x^2 - 4}{x^2 - 16})}\right) = \frac{\pi}{4}. \]

Since \(\tan\left(\frac{\pi}{4}\right) = 1\), we set the argument of the tangent equal to 1:

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

### Step 6: Solve the equation

Cross-multiplying gives:

\[ 2x^2 - 16 = (x-4)(x+4)\left(1 - \frac{x^2 - 4}{x^2 - 16}\right). \]

The left-hand side simplifies to:

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

This leads to:

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

Simplifying, we find:

\[ 0 = -x^2 + 4 \implies x^2 = 4 \implies x = \pm 2. \]

### Step 7: Verify solutions

We need to check if both solutions satisfy the original equation.

1. For \(x = 2\):
\[ \tan^{-1}\left(\frac{0}{-2}\right) + \tan^{-1}\left(\frac{4}{6}\right) \neq \frac{\pi}{4}. \]

2. For \(x = -2\):
\[ \tan^{-1}\left(\frac{-4}{-6}\right) + \tan^{-1}\left(\frac{0}{2}\right) = \frac{\pi}{4}. \]

Thus, the only valid solution is:

\[ \boxed{-2}. \]