If `"tan"^(-1)x+"tan"^(-1)"1"/x={(pi//k,if", "xge0),(-pi//k,if", "xlt0):}` , then the value of K is

Options

A. 3
B. 4
C. 2
D. 1

Answer

To solve the problem, we need to find the value of \( k \) in the equation:

\[ \tan^{-1} x + \tan^{-1} \frac{1}{x} = \begin{cases} \frac{\pi}{k} & \text{if } x \geq 0 \\ -\frac{\pi}{k} & \text{if } x < 0 \end{cases} \]

### Step 1: Analyze the case when \( x \geq 0 \)

For \( x \geq 0 \):
- We know that \( \tan^{-1} \frac{1}{x} = \cot^{-1} x \) because \( \tan^{-1} \) and \( \cot^{-1} \) are complementary functions when \( x > 0 \).
- Thus, we can write:

\[ \tan^{-1} x + \tan^{-1} \frac{1}{x} = \tan^{-1} x + \cot^{-1} x \]

### Step 2: Use the identity for \( \tan^{-1} x + \cot^{-1} x \)

From trigonometric identities, we know:

\[ \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \]

So, for \( x \geq 0 \):

\[ \tan^{-1} x + \tan^{-1} \frac{1}{x} = \frac{\pi}{2} \]

### Step 3: Analyze the case when \( x < 0 \)

For \( x < 0 \):
- We cannot directly use \( \cot^{-1} x \) because it is not defined in the same way for negative values. Instead, we can express \( \tan^{-1} \frac{1}{x} \) as:

\[ \tan^{-1} \frac{1}{x} = \cot^{-1} x - \pi \]

Thus, we have:

\[ \tan^{-1} x + \tan^{-1} \frac{1}{x} = \tan^{-1} x + \left(\cot^{-1} x - \pi\right) \]

### Step 4: Use the identity for \( \tan^{-1} x + \cot^{-1} x \) again

Using the same identity as before:

\[ \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \]

So, for \( x < 0 \):

\[ \tan^{-1} x + \tan^{-1} \frac{1}{x} = \frac{\pi}{2} - \pi = -\frac{\pi}{2} \]

### Step 5: Set up the equations

Now we have:

\[ \frac{\pi}{2} = \frac{\pi}{k} \quad \text{(for } x \geq 0\text{)} \]
\[ -\frac{\pi}{2} = -\frac{\pi}{k} \quad \text{(for } x < 0\text{)} \]

### Step 6: Solve for \( k \)

From the first equation:

\[ \frac{\pi}{2} = \frac{\pi}{k} \implies k = 2 \]

From the second equation:

\[ -\frac{\pi}{2} = -\frac{\pi}{k} \implies k = 2 \]

### Conclusion

In both cases, we find that \( k = 2 \).

Thus, the value of \( k \) is:

\[ \boxed{2} \]