Draw the graph of `y=(sin2x)sqrt(1+tan^2x)` , find its domain and range.

Explanation

To solve the problem of drawing the graph of \( y = \sin(2x) \sqrt{1 + \tan^2(x)} \) and finding its domain and range, we will follow these steps:

### Step 1: Rewrite the Function
We start with the function:
\[ y = \sin(2x) \sqrt{1 + \tan^2(x)} \]
We can use the identity \( \sqrt{1 + \tan^2(x)} = \sec(x) \) to rewrite the function:
\[ y = \sin(2x) \sec(x) \]

### Step 2: Express \(\sin(2x)\)
Using the double angle identity for sine, we can express \(\sin(2x)\) as:
\[ \sin(2x) = 2 \sin(x) \cos(x) \]
Thus, we can rewrite \(y\) as:
\[ y = 2 \sin(x) \cos(x) \sec(x) \]
Since \(\sec(x) = \frac{1}{\cos(x)}\), we have:
\[ y = 2 \sin(x) \]

### Step 3: Determine the Domain
The function \(y = 2 \sin(x)\) is defined for all \(x\) except where \(\cos(x) = 0\) (since \(\sec(x)\) is undefined). The values of \(x\) where \(\cos(x) = 0\) are:
\[ x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]
Thus, the domain of \(y\) is:
\[ \text{Domain: } x \in \mathbb{R}, x \neq \frac{\pi}{2} + n\pi \]

### Step 4: Determine the Range
The function \(y = 2 \sin(x)\) oscillates between -2 and 2, as the sine function oscillates between -1 and 1. Therefore, the range of \(y\) is:
\[ \text{Range: } [-2, 2] \]

### Step 5: Sketch the Graph
To sketch the graph of \(y = 2 \sin(x)\):
1. The graph will oscillate between -2 and 2.
2. It will have zeros at \(x = n\pi\) for \(n \in \mathbb{Z}\).
3. The maximum value of 2 occurs at \(x = \frac{\pi}{2} + 2n\pi\) and the minimum value of -2 occurs at \(x = \frac{3\pi}{2} + 2n\pi\).

### Final Summary
- **Domain**: \( x \in \mathbb{R}, x \neq \frac{\pi}{2} + n\pi \)
- **Range**: \([-2, 2]\)