If `xlt0`, then value of `tan^(-1)(x)+tan^(-1)(1/x)` is equal to

Options

A. `(pi)/2`
B. `-(pi)/2`
C. `0`
D. `-pi`

Explanation

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

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

### Step 2: Apply the identity
Let \( a = x \) and \( b = \frac{1}{x} \). Then we have:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}}\right) \]

### Step 3: Simplify the expression
Now, simplify the expression:
\[ 1 - x \cdot \frac{1}{x} = 1 - 1 = 0 \]
Thus, the denominator becomes zero.

### Step 4: Analyze the result
Since the denominator is zero, the expression becomes undefined. However, we know that when the denominator of the tangent function approaches zero, the value of the tangent function approaches \( \pm \infty \).

### Step 5: Determine the angle
Since \( x < 0 \), we need to determine the angle:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) \to \text{undefined} \text{ at } \frac{\pi}{2} \text{ or } -\frac{\pi}{2} \]
Given that \( x < 0 \), we conclude that:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = -\frac{\pi}{2} \]

### Final Answer
Thus, the value of \( \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) \) is:
\[ -\frac{\pi}{2} \]

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