Express `1/(1+cos theta-i sin theta)` in the form of `a +ib`.

Answer

To express \( \frac{1}{1 + \cos \theta - i \sin \theta} \) in the form \( a + ib \), we will follow these steps:

### Step 1: Rationalize the Denominator
To eliminate the imaginary unit \( i \) from the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is \( 1 + \cos \theta + i \sin \theta \).

\[ \frac{1}{1 + \cos \theta - i \sin \theta} \cdot \frac{1 + \cos \theta + i \sin \theta}{1 + \cos \theta + i \sin \theta} \]

### Step 2: Simplify the Numerator
The numerator becomes:

\[ 1 + \cos \theta + i \sin \theta \]

### Step 3: Simplify the Denominator
The denominator can be simplified using the identity \( (a - b)(a + b) = a^2 - b^2 \), where \( a = 1 + \cos \theta \) and \( b = i \sin \theta \):

\[ (1 + \cos \theta)^2 - (i \sin \theta)^2 \]

Calculating each part:

\[ (1 + \cos \theta)^2 = 1 + 2\cos \theta + \cos^2 \theta \]
\[ (i \sin \theta)^2 = -\sin^2 \theta \]

Thus, the denominator becomes:

\[ 1 + 2\cos \theta + \cos^2 \theta + \sin^2 \theta \]

### Step 4: Use the Pythagorean Identity
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ 1 + 2\cos \theta + 1 = 2 + 2\cos \theta \]

### Step 5: Combine the Results
Now we have:

\[ \frac{1 + \cos \theta + i \sin \theta}{2 + 2\cos \theta} \]

### Step 6: Separate Real and Imaginary Parts
We can express this as:

\[ \frac{1 + \cos \theta}{2 + 2\cos \theta} + i \frac{\sin \theta}{2 + 2\cos \theta} \]

### Final Result
Thus, we can express \( \frac{1}{1 + \cos \theta - i \sin \theta} \) in the form \( a + ib \):

\[ a = \frac{1 + \cos \theta}{2 + 2\cos \theta}, \quad b = \frac{\sin \theta}{2 + 2\cos \theta} \]