Solve the equation `(1-tantheta)(1+sin2theta)=1+tantheta`

Answer

To solve the equation \( (1 - \tan \theta)(1 + \sin 2\theta) = 1 + \tan \theta \), we will follow these steps:

### Step 1: Expand the left-hand side
We start with the equation:
\[ (1 - \tan \theta)(1 + \sin 2\theta) = 1 + \tan \theta \]
Expanding the left-hand side gives:
\[ 1 + \sin 2\theta - \tan \theta - \tan \theta \sin 2\theta = 1 + \tan \theta \]

### Step 2: Simplify the equation
Next, we can cancel the \(1\) from both sides:
\[ \sin 2\theta - \tan \theta - \tan \theta \sin 2\theta = \tan \theta \]
Rearranging this gives:
\[ \sin 2\theta - \tan \theta \sin 2\theta = 2\tan \theta \]

### Step 3: Substitute \(\sin 2\theta\)
Recall that \(\sin 2\theta = 2 \sin \theta \cos \theta\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Substitute these into the equation:
\[ 2 \sin \theta \cos \theta - \left(\frac{\sin \theta}{\cos \theta}\right)(2 \sin \theta \cos \theta) = 2\left(\frac{\sin \theta}{\cos \theta}\right) \]
This simplifies to:
\[ 2 \sin \theta \cos \theta - 2 \sin^2 \theta = \frac{2 \sin \theta}{\cos \theta} \]

### Step 4: Multiply through by \(\cos \theta\)
To eliminate the fraction, multiply the entire equation by \(\cos \theta\):
\[ 2 \sin \theta \cos^2 \theta - 2 \sin^2 \theta \cos \theta = 2 \sin \theta \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 2 \sin \theta \cos^2 \theta - 2 \sin^2 \theta \cos \theta - 2 \sin \theta = 0 \]
Factoring out \(2 \sin \theta\):
\[ 2 \sin \theta (\cos^2 \theta - \sin \theta \cos \theta - 1) = 0 \]

### Step 6: Solve for \(\sin \theta = 0\)
From \(2 \sin \theta = 0\), we have:
\[ \sin \theta = 0 \implies \theta = n\pi, \quad n \in \mathbb{Z} \]

### Step 7: Solve the quadratic equation
Now we solve the quadratic equation:
\[ \cos^2 \theta - \sin \theta \cos \theta - 1 = 0 \]
Using the substitution \(\sin \theta = \sqrt{1 - \cos^2 \theta}\), we can solve for \(\theta\) in terms of \(\cos \theta\).

### Step 8: Find the roots
Using the quadratic formula:
\[ \cos \theta = \frac{\sin \theta \pm \sqrt{(\sin \theta)^2 + 4}}{2} \]
We can find the values of \(\theta\).

### Final Solution
Thus, the solutions to the equation \( (1 - \tan \theta)(1 + \sin 2\theta) = 1 + \tan \theta \) are:
\[ \theta = n\pi, \quad n \in \mathbb{Z} \]
and any additional solutions from the quadratic equation.