If `y=tan^(-1)x+tan^(-1)(1/x)+sec^(-1)x ,` then `y` lies in the interval ` (a) [pi/2,pi)uu[pi,(3pi)/2)` (b) `[pi/2,(3pi)/2]` `(c)(0,pi)` (d) `[0,pi/2)uu[pi/2,pi)`

Answer

To solve the problem, we need to analyze the expression given:

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

### Step 1: Understanding the components of y

1. **Domain of \(\tan^{-1} x\)**:
- The function \(\tan^{-1} x\) is defined for all real numbers \(x\) and its range is \((- \frac{\pi}{2}, \frac{\pi}{2})\).

2. **Domain of \(\tan^{-1} \left(\frac{1}{x}\right)\)**:
- This function is defined for \(x \neq 0\) and its range is also \((- \frac{\pi}{2}, \frac{\pi}{2})\).

3. **Domain of \(\sec^{-1} x\)**:
- The function \(\sec^{-1} x\) is defined for \(x \leq -1\) or \(x \geq 1\). Its range is \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\).

### Step 2: Finding the overall domain of y

- The overall domain of \(y\) is the intersection of the domains of the individual components:
- \(\tan^{-1} x\) and \(\tan^{-1} \left(\frac{1}{x}\right)\) restrict \(x\) to \((- \frac{\pi}{2}, \frac{\pi}{2})\) excluding \(0\).
- \(\sec^{-1} x\) restricts \(x\) to \(x \leq -1\) or \(x \geq 1\).

Thus, the intersection of these domains gives us:
- For \(x \geq 1\): \((- \frac{\pi}{2}, \frac{\pi}{2}) \cap [1, \infty) = [1, \infty)\)
- For \(x \leq -1\): \((- \frac{\pi}{2}, \frac{\pi}{2}) \cap (-\infty, -1] = (-\infty, -1)\)

### Step 3: Evaluating y at the boundaries

1. **As \(x \to 1\)**:
\[ y = \tan^{-1}(1) + \tan^{-1}(1) + \sec^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{4} + 0 = \frac{\pi}{2} \]

2. **As \(x \to -1\)**:
\[ y = \tan^{-1}(-1) + \tan^{-1}(-1) + \sec^{-1}(-1) = -\frac{\pi}{4} - \frac{\pi}{4} + \pi = \frac{\pi}{2} \]

3. **As \(x \to \infty\)**:
\[ y = \tan^{-1}(\infty) + \tan^{-1}(0) + \sec^{-1}(\infty) = \frac{\pi}{2} + 0 + 0 = \frac{\pi}{2} \]

4. **As \(x \to -\infty\)**:
\[ y = \tan^{-1}(-\infty) + \tan^{-1}(0) + \sec^{-1}(-\infty) = -\frac{\pi}{2} + 0 + \pi = \frac{\pi}{2} \]

### Step 4: Conclusion on the range of y

From the evaluations, we see that \(y\) approaches \(\frac{\pi}{2}\) but does not include it since both \(\tan^{-1}\) functions are exclusive at their limits. Therefore, the range of \(y\) is:

\[ y \in (0, \pi) \]

### Final Answer

Thus, the correct option is:
**(c) (0, π)**