If `x lt 0` then `tan^(-1)(1/x)` equals

Options

A. `cot^(-1)x`
B. `-cot^(-1)x`
C. `-pi + cot^(-1)x`
D. `-pi -cot^(-1) x `

Answer

To solve the problem, we need to find the value of \( \tan^{-1}\left(\frac{1}{x}\right) \) given that \( x < 0 \).

### Step-by-Step Solution:

1. **Understanding the Function**:
We start with the expression \( \tan^{-1}\left(\frac{1}{x}\right) \). Since \( x < 0 \), \( \frac{1}{x} \) will also be negative.

2. **Using the Identity**:
We can use the identity that relates the tangent and cotangent functions:
\[ \tan^{-1}(y) = \cot^{-1}\left(\frac{1}{y}\right) \]
Therefore, we can write:
\[ \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}(x) \]

3. **Considering the Sign**:
Since \( x < 0 \), we have to consider the properties of the inverse cotangent function. The cotangent function is defined as:
\[ \cot^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(y) \]
However, for negative values, we can also express it as:
\[ \cot^{-1}(-y) = \pi - \cot^{-1}(y) \]
Thus, we can write:
\[ \cot^{-1}(x) = \pi - \cot^{-1}(-x) \]

4. **Final Expression**:
Since \( x < 0 \), we can express \( \tan^{-1}\left(\frac{1}{x}\right) \) in terms of \( \cot^{-1}(x) \):
\[ \tan^{-1}\left(\frac{1}{x}\right) = \pi - \cot^{-1}(-x) \]
But since \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \), we can simplify:
\[ \tan^{-1}\left(\frac{1}{x}\right) = -\frac{\pi}{2} + \cot^{-1}(x) \]

5. **Conclusion**:
Therefore, the final result is:
\[ \tan^{-1}\left(\frac{1}{x}\right) = -\frac{\pi}{2} + \cot^{-1}(x) \]