If `tan^(-1)((x-1)/(x-2))+cot^(-1)((x+2)/(x+1))=(pi)/(4)` , find x.

Answer

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

### Step 1: Rewrite the cotangent inverse
We know that \( \cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right) \). Therefore, we can rewrite the equation as:
\[ \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{1}{\frac{x+2}{x+1}}\right) = \frac{\pi}{4} \]
This simplifies to:
\[ \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{x+1}{x+2}\right) = \frac{\pi}{4} \]

### Step 2: Use the tangent addition formula
Using the formula \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \) when \( ab < 1 \), we can set:
\[ a = \frac{x-1}{x-2}, \quad b = \frac{x+1}{x+2} \]
Thus, we have:
\[ \tan^{-1}\left(\frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right) = \frac{\pi}{4} \]
This implies:
\[ \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \frac{x-1}{x-2} \cdot \frac{x+1}{x+2}} = 1 \]

### Step 3: Simplify the expression
Now, we need to simplify the numerator and denominator:
1. **Numerator**:
\[ \frac{x-1}{x-2} + \frac{x+1}{x+2} = \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)} \]
Expanding this gives:
\[ (x^2 + 2x - x - 2) + (x^2 - 2x + x - 2) = 2x^2 - 4 \]
So, the numerator becomes:
\[ \frac{2x^2 - 4}{(x-2)(x+2)} \]

2. **Denominator**:
\[ 1 - \frac{(x-1)(x+1)}{(x-2)(x+2)} = \frac{(x-2)(x+2) - (x^2 - 1)}{(x-2)(x+2)} \]
Simplifying gives:
\[ \frac{x^2 - 4 - x^2 + 1}{(x-2)(x+2)} = \frac{-3}{(x-2)(x+2)} \]

### Step 4: Set up the equation
Now we can set up the equation:
\[ \frac{2x^2 - 4}{(x-2)(x+2)} \cdot \frac{(x-2)(x+2)}{-3} = 1 \]
This simplifies to:
\[ \frac{2x^2 - 4}{-3} = 1 \]
Multiplying both sides by -3 gives:
\[ 2x^2 - 4 = -3 \]
Thus:
\[ 2x^2 = 1 \quad \Rightarrow \quad x^2 = \frac{1}{2} \quad \Rightarrow \quad x = \pm \frac{1}{\sqrt{2}} \]

### Final Answer
Therefore, the values of \( x \) are:
\[ x = \frac{1}{\sqrt{2}} \quad \text{or} \quad x = -\frac{1}{\sqrt{2}} \]