If `(-pi)/(2) le x le (3pi)/(2)` then tanx is an increasing function. Check whether True or False.

Answer

To determine whether the statement "tan x is an increasing function on the interval \([- \frac{\pi}{2}, \frac{3\pi}{2}]\)" is true or false, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \tan x \).

### Step 2: Find the derivative
To check if \( f(x) \) is an increasing function, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x \]

### Step 3: Analyze the derivative
The function \( \sec^2 x \) is always positive wherever it is defined. The secant function, \( \sec x \), is defined wherever \( \cos x \neq 0 \).

### Step 4: Determine the points of discontinuity
The function \( \tan x \) has vertical asymptotes (and thus is undefined) at:
\[ x = -\frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]
Within the interval \([- \frac{\pi}{2}, \frac{3\pi}{2}]\), the points where \( \tan x \) is undefined are:
- \( x = -\frac{\pi}{2} \) (left endpoint)
- \( x = \frac{\pi}{2} \) (discontinuity)
- \( x = \frac{3\pi}{2} \) (right endpoint)

### Step 5: Analyze the intervals
We will analyze the behavior of \( \tan x \) in the intervals:
1. \([- \frac{\pi}{2}, \frac{\pi}{2})\)
2. \((\frac{\pi}{2}, \frac{3\pi}{2})\)

- In the interval \([- \frac{\pi}{2}, \frac{\pi}{2})\), \( \tan x \) is increasing because \( f'(x) = \sec^2 x > 0 \).
- In the interval \((\frac{\pi}{2}, \frac{3\pi}{2})\), \( \tan x \) is also increasing because \( f'(x) = \sec^2 x > 0 \).

### Step 6: Conclusion
Since \( f'(x) > 0 \) in both intervals where \( \tan x \) is defined, we conclude that \( \tan x \) is indeed an increasing function on the interval \([- \frac{\pi}{2}, \frac{3\pi}{2}]\) (excluding the points where it is undefined).

Thus, the statement is **True**.