Solve the following equations :
` tan^(-1) (x+2) +tan^(-1) (x-2) = pi/4 , x gt 0 `

Answer

To solve the equation \( \tan^{-1}(x+2) + \tan^{-1}(x-2) = \frac{\pi}{4} \) for \( x > 0 \), we can follow these steps:

### Step 1: Use the formula for the sum of inverse tangents
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
provided \( ab < 1 \).

In our case, let \( a = x + 2 \) and \( b = x - 2 \). Thus, we can rewrite the equation as:
\[ \tan^{-1}\left(\frac{(x + 2) + (x - 2)}{1 - (x + 2)(x - 2)}\right) = \frac{\pi}{4} \]

### Step 2: Simplify the expression
Now, simplify the numerator and denominator:
- Numerator:
\[ (x + 2) + (x - 2) = 2x \]
- Denominator:
\[ 1 - (x + 2)(x - 2) = 1 - (x^2 - 4) = 5 - x^2 \]

So, we have:
\[ \tan^{-1}\left(\frac{2x}{5 - x^2}\right) = \frac{\pi}{4} \]

### Step 3: Take the tangent of both sides
Taking the tangent of both sides gives:
\[ \frac{2x}{5 - x^2} = 1 \]

### Step 4: Solve for \( x \)
Cross-multiplying leads to:
\[ 2x = 5 - x^2 \]
Rearranging gives:
\[ x^2 + 2x - 5 = 0 \]

### Step 5: Use the quadratic formula
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 2, c = -5 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6} \]

### Step 6: Determine valid solutions
This gives us two potential solutions:
\[ x = -1 + \sqrt{6} \quad \text{and} \quad x = -1 - \sqrt{6} \]
Since \( x > 0 \), we only consider \( x = -1 + \sqrt{6} \).

### Final Answer
Thus, the solution to the equation is:
\[ x = -1 + \sqrt{6} \]