`tan^(- 1)(x+2/x)-tan^(- 1)(4/x)=tan^(- 1)(x-2/x)`

Options

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

Explanation

To solve the equation

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

we will use the formula for the difference of two inverse tangents:

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

### Step 1: Apply the formula

Let \( a = \frac{x+2}{x} \) and \( b = \frac{4}{x} \). Then, we can rewrite the left-hand side:

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

### Step 2: Simplify \( a - b \)

Calculating \( a - b \):

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

### Step 3: Simplify \( 1 + ab \)

Calculating \( ab \):

\[ ab = \left(\frac{x+2}{x}\right)\left(\frac{4}{x}\right) = \frac{4(x+2)}{x^2} = \frac{4x + 8}{x^2}. \]

Thus,

\[ 1 + ab = 1 + \frac{4x + 8}{x^2} = \frac{x^2 + 4x + 8}{x^2}. \]

### Step 4: Combine results

Now substituting back into the formula:

\[ \tan^{-1}\left(\frac{\frac{x-2}{x}}{\frac{x^2 + 4x + 8}{x^2}}\right) = \tan^{-1}\left(\frac{(x-2)x^2}{x(x^2 + 4x + 8)}\right) = \tan^{-1}\left(\frac{x^2(x-2)}{x^2 + 4x + 8}\right). \]

### Step 5: Set equal to the right-hand side

Now we have:

\[ \tan^{-1}\left(\frac{x^2(x-2)}{x^2 + 4x + 8}\right) = \tan^{-1}\left(\frac{x-2}{x}\right). \]

### Step 6: Equate the arguments

Since the inverse tangent function is one-to-one, we can equate the arguments:

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

### Step 7: Cross-multiply

Cross-multiplying gives:

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

### Step 8: Simplify

Assuming \( x - 2 \neq 0 \) (we will check this case later), we can divide both sides by \( x - 2 \):

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

### Step 9: Rearrange

Rearranging gives:

\[ x^3 - 2x^2 - x^2 - 4x - 8 = 0 \implies x^3 - 3x^2 - 4x - 8 = 0. \]

### Step 10: Factor or use the Rational Root Theorem

Using the Rational Root Theorem or synthetic division, we can find that \( x = 4 \) is a root. Dividing \( x^3 - 3x^2 - 4x - 8 \) by \( x - 4 \) gives:

\[ x^2 + x + 2 = 0. \]

### Step 11: Solve the quadratic

The quadratic \( x^2 + x + 2 = 0 \) has no real roots (discriminant \( 1 - 8 < 0 \)). Thus, the only real solution is \( x = 4 \).

### Step 12: Check for \( x = 2 \)

We also need to check if \( x - 2 = 0 \) leads to a solution:

Setting \( x = 2 \):

\[ \tan^{-1}(0) - \tan^{-1}(2) = \tan^{-1}(0) \implies -\tan^{-1}(2) = 0, \]

which is not true.

### Final Solution

Thus, the only solution is:

\[ \boxed{4}. \]