The real part of `( 1- cos theta + 2 i sin theta )^(-1)` is

Options

A. `(1)/( 5+ 3 cos theta )`
B. `( 1)/( 5- 3 cos theta )`
C. `( 1)/( 3-5 cos theta )`
D. `( 1)/( 5-3 cos theta )`

Answer

To find the real part of the expression \( (1 - \cos \theta + 2i \sin \theta)^{-1} \), we can follow these steps:

### Step 1: Write the expression in a more manageable form.
We start with the expression:
\[ z = 1 - \cos \theta + 2i \sin \theta \]
We need to find \( z^{-1} \).

### Step 2: Rationalize the denominator.
To find \( z^{-1} \), we multiply the numerator and denominator by the conjugate of the denominator:
\[ z^{-1} = \frac{1 - \cos \theta - 2i \sin \theta}{(1 - \cos \theta + 2i \sin \theta)(1 - \cos \theta - 2i \sin \theta)} \]

### Step 3: Simplify the denominator.
Now we calculate the denominator:
\[ (1 - \cos \theta)^2 + (2 \sin \theta)^2 \]
Expanding this:
\[ (1 - \cos \theta)^2 = 1 - 2\cos \theta + \cos^2 \theta \]
\[ (2 \sin \theta)^2 = 4 \sin^2 \theta \]
So the denominator becomes:
\[ 1 - 2\cos \theta + \cos^2 \theta + 4 \sin^2 \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = 1 - 2\cos \theta + \cos^2 \theta + 4(1 - \cos^2 \theta) \]
\[ = 1 - 2\cos \theta + \cos^2 \theta + 4 - 4\cos^2 \theta \]
\[ = 5 - 2\cos \theta - 3\cos^2 \theta \]

### Step 4: Write the complete expression for \( z^{-1} \).
Now we can write:
\[ z^{-1} = \frac{1 - \cos \theta - 2i \sin \theta}{5 - 2\cos \theta - 3\cos^2 \theta} \]

### Step 5: Separate the real and imaginary parts.
The real part of \( z^{-1} \) is:
\[ \text{Re}(z^{-1}) = \frac{1 - \cos \theta}{5 - 2\cos \theta - 3\cos^2 \theta} \]

### Step 6: Final expression for the real part.
Thus, the real part of \( (1 - \cos \theta + 2i \sin \theta)^{-1} \) is:
\[ \text{Re}(z^{-1}) = \frac{1 - \cos \theta}{5 + 3\cos \theta} \]