The number of roots of `(1-tan theta) (1+sin 2 theta)=1+tan theta` for `theta in [0, 2pi]` is

Options

A. 3
B. 4
C. 5
D. none of these

Explanation

To find the number of roots of the equation \((1 - \tan \theta)(1 + \sin 2\theta) = 1 + \tan \theta\) for \(\theta\) in the interval \([0, 2\pi]\), we will solve the equation step by step.

### Step 1: Rewrite the equation
We start with the given equation:
\[ (1 - \tan \theta)(1 + \sin 2\theta) = 1 + \tan \theta \]
Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta = \frac{2\tan \theta}{1 + \tan^2 \theta}\), we can rewrite \(\sin 2\theta\):
\[ (1 - \tan \theta)\left(1 + \frac{2\tan \theta}{1 + \tan^2 \theta}\right) = 1 + \tan \theta \]

### Step 2: Simplify the left-hand side
Now we simplify the left-hand side:
\[ (1 - \tan \theta)\left(\frac{1 + \tan^2 \theta + 2\tan \theta}{1 + \tan^2 \theta}\right) = 1 + \tan \theta \]
This simplifies to:
\[ \frac{(1 - \tan \theta)(1 + \tan^2 \theta + 2\tan \theta)}{1 + \tan^2 \theta} = 1 + \tan \theta \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (1 - \tan \theta)(1 + \tan^2 \theta + 2\tan \theta) = (1 + \tan \theta)(1 + \tan^2 \theta) \]

### Step 4: Expand both sides
Expanding both sides:
Left-hand side:
\[ 1 + \tan^2 \theta + 2\tan \theta - \tan \theta - \tan^3 \theta - 2\tan^2 \theta = 1 + \tan^2 \theta \]
Right-hand side:
\[ 1 + \tan \theta + \tan^2 \theta + \tan^3 \theta \]

### Step 5: Set the equation to zero
Setting the equation to zero:
\[ 1 + \tan^2 \theta + \tan \theta - \tan^3 \theta - 2\tan^2 \theta = 1 + \tan \theta + \tan^2 \theta + \tan^3 \theta \]
This simplifies to:
\[ -\tan^3 \theta - \tan^2 \theta - 2\tan^2 \theta = 0 \]
\[ -\tan^3 \theta - 3\tan^2 \theta = 0 \]
Factoring out \(-\tan^2 \theta\):
\[ -\tan^2 \theta(\tan \theta + 3) = 0 \]

### Step 6: Solve for \(\tan \theta\)
Setting each factor to zero gives:
1. \(-\tan^2 \theta = 0 \Rightarrow \tan \theta = 0\)
2. \(\tan \theta + 3 = 0 \Rightarrow \tan \theta = -3\)

### Step 7: Find the angles
For \(\tan \theta = 0\):
\[ \theta = n\pi \quad \text{for } n \in \mathbb{Z} \]
In the interval \([0, 2\pi]\), the solutions are:
\[ \theta = 0, \pi \]

For \(\tan \theta = -3\):
\[ \theta = \tan^{-1}(-3) + n\pi \]
Calculating the angles in the interval \([0, 2\pi]\):
- The principal value \(\tan^{-1}(-3)\) gives an angle in the fourth quadrant, which can be adjusted to find the corresponding angles in the specified interval.

### Step 8: Count the roots
Thus, we have:
1. \(\theta = 0\)
2. \(\theta = \pi\)
3. Two angles corresponding to \(\tan \theta = -3\) in the intervals.

In total, we find **4 roots** in the interval \([0, 2\pi]\).

### Final Answer
The number of roots of the equation is **4**.