`(1+sin 2theta+cos 2theta)/(1+sin2 theta-cos 2 theta)`=

Options

A. `(1)/(2)tan theta`
B. `(1)/(2)cot theta`
C. `tan theta`
D. `cot theta`

Answer

To solve the expression \(\frac{1 + \sin 2\theta + \cos 2\theta}{1 + \sin 2\theta - \cos 2\theta}\), we can follow these steps:

### Step-by-step Solution:

1. **Recall the double angle identities**:
- We know that \(\sin 2\theta = 2 \sin \theta \cos \theta\)
- We also have two forms for \(\cos 2\theta\):
\[ \cos 2\theta = 2 \cos^2 \theta - 1 \quad \text{or} \quad \cos 2\theta = 1 - 2 \sin^2 \theta \]

2. **Substituting \(\cos 2\theta\)**:
- Using \(\cos 2\theta = 2 \cos^2 \theta - 1\), we can rewrite the expression:
\[ 1 + \sin 2\theta + \cos 2\theta = 1 + 2 \sin \theta \cos \theta + (2 \cos^2 \theta - 1) \]
- Simplifying this gives:
\[ 2 \sin \theta \cos \theta + 2 \cos^2 \theta = 2 \cos \theta (\sin \theta + \cos \theta) \]

3. **Substituting \(\cos 2\theta\) in the denominator**:
- Now for the denominator:
\[ 1 + \sin 2\theta - \cos 2\theta = 1 + 2 \sin \theta \cos \theta - (2 \cos^2 \theta - 1) \]
- Simplifying this gives:
\[ 2 \sin \theta \cos \theta + 1 - 2 \cos^2 \theta = 2 \sin \theta \cos \theta + 2 \sin^2 \theta = 2 \sin \theta (\sin \theta + \cos \theta) \]

4. **Putting it all together**:
- Now substituting the simplified forms back into the original expression:
\[ \frac{2 \cos \theta (\sin \theta + \cos \theta)}{2 \sin \theta (\sin \theta + \cos \theta)} \]
- The \((\sin \theta + \cos \theta)\) terms cancel out (assuming \(\sin \theta + \cos \theta \neq 0\)):
\[ \frac{2 \cos \theta}{2 \sin \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta \]

5. **Final Result**:
- Therefore, the final answer is:
\[ \cot \theta \]