If `y=tan^(-1)x+tan^(-1)(1/x)+cosec^(-1)x,xepsilon(-oo,-1)uu[1,oo)`, then `yepsilon`

Options

A. `[-pi,-(pi)/2)uu((pi)/2,pi]`
B. `(-(pi)/2,(pi)/2)`
C. `(0,pi)`
D. `[0,(pi)/2)uu[(pi)/2,pi)`

Explanation

To solve the problem, we need to analyze the given expression for \( y \):

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

where \( x \in (-\infty, -1) \cup [1, \infty) \).

### Step 1: Simplifying \( \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) \)

Using the identity for the sum of arctangents, we have:

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \quad \text{for } x > 0 \]
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = -\frac{\pi}{2} \quad \text{for } x < 0 \]

Since \( x \) can be either negative or positive, we will consider both cases.

### Step 2: Case 1: \( x \in [1, \infty) \)

For \( x \geq 1 \):

\[ y = \frac{\pi}{2} + \csc^{-1}(x) \]

The range of \( \csc^{-1}(x) \) for \( x \geq 1 \) is \( [0, \frac{\pi}{2}] \). Therefore, we can find the range of \( y \):

\[ y \in \left[\frac{\pi}{2}, \frac{\pi}{2} + \frac{\pi}{2}\right] = \left[\frac{\pi}{2}, \pi\right] \]

### Step 3: Case 2: \( x \in (-\infty, -1) \)

For \( x < -1 \):

\[ y = -\frac{\pi}{2} + \csc^{-1}(x) \]

The range of \( \csc^{-1}(x) \) for \( x < -1 \) is \( (-\pi, -\frac{\pi}{2}] \). Therefore, we can find the range of \( y \):

\[ y \in \left[-\frac{\pi}{2} - \pi, -\frac{\pi}{2}\right] = \left[-\frac{3\pi}{2}, -\frac{\pi}{2}\right] \]

### Step 4: Combining the Ranges

Now we combine the ranges from both cases:

1. For \( x \in [1, \infty) \), \( y \in \left[\frac{\pi}{2}, \pi\right] \)
2. For \( x \in (-\infty, -1) \), \( y \in \left[-\frac{3\pi}{2}, -\frac{\pi}{2}\right] \)

Thus, the overall range of \( y \) is:

\[ y \in \left[-\frac{3\pi}{2}, -\frac{\pi}{2}\right] \cup \left[\frac{\pi}{2}, \pi\right] \]

### Final Answer

The final answer is:

\[ y \in \left[-\frac{3\pi}{2}, -\frac{\pi}{2}\right] \cup \left[\frac{\pi}{2}, \pi\right] \]