If `tan^-1(1/y)=-pi+cot^-1 y,` where `y=x^2-3x+2,` then find the value of `x`

Explanation

To solve the equation \( \tan^{-1}\left(\frac{1}{y}\right) = -\pi + \cot^{-1}(y) \), where \( y = x^2 - 3x + 2 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan^{-1}\left(\frac{1}{y}\right) = -\pi + \cot^{-1}(y) \]

### Step 2: Use the identity for cotangent
Recall that \( \cot^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(y) \). Thus, we can rewrite the right-hand side:
\[ -\pi + \cot^{-1}(y) = -\pi + \left(\frac{\pi}{2} - \tan^{-1}(y)\right) = -\frac{3\pi}{2} - \tan^{-1}(y) \]

### Step 3: Set the equations equal
Now we have:
\[ \tan^{-1}\left(\frac{1}{y}\right) = -\frac{3\pi}{2} - \tan^{-1}(y) \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \tan^{-1}\left(\frac{1}{y}\right) + \tan^{-1}(y) = -\frac{3\pi}{2} \]

### Step 5: Analyze the conditions
For the equation \( \tan^{-1}\left(\frac{1}{y}\right) \) to be defined and equal to a negative angle, \( y \) must be negative. Thus, we must have:
\[ y < 0 \]

### Step 6: Substitute for \( y \)
Substituting \( y = x^2 - 3x + 2 \):
\[ x^2 - 3x + 2 < 0 \]

### Step 7: Factor the quadratic
Factoring the quadratic gives:
\[ (x - 1)(x - 2) < 0 \]

### Step 8: Determine the intervals
The roots of the equation are \( x = 1 \) and \( x = 2 \). The quadratic will be negative between the roots:
\[ 1 < x < 2 \]

### Step 9: Conclusion
Thus, the solution for \( x \) is:
\[ x \in (1, 2) \]