Express the complex number in the form `r(cos theta+isin theta)`
(ii)`1-sin alpha +icos alpha`

Answer

To express the complex number \(1 - \sin \alpha + i \cos \alpha\) in the form \(r(\cos \theta + i \sin \theta)\), we will follow these steps:

### Step 1: Identify \(x\) and \(y\)
In the complex number \(1 - \sin \alpha + i \cos \alpha\), we can identify:
- \(x = 1 - \sin \alpha\)
- \(y = \cos \alpha\)

### Step 2: Calculate \(r\)
The modulus \(r\) of the complex number is given by the formula:
\[ r = \sqrt{x^2 + y^2} \]
Substituting the values of \(x\) and \(y\):
\[ r = \sqrt{(1 - \sin \alpha)^2 + (\cos \alpha)^2} \]

### Step 3: Simplify \(r\)
Now, we will simplify \(r\):
\[ r = \sqrt{(1 - 2\sin \alpha + \sin^2 \alpha) + \cos^2 \alpha} \]
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):
\[ r = \sqrt{1 - 2\sin \alpha + 1} = \sqrt{2(1 - \sin \alpha)} \]

### Step 4: Calculate \(\theta\)
To find \(\theta\), we use the formula:
\[ \tan \theta = \frac{y}{x} = \frac{\cos \alpha}{1 - \sin \alpha} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{\cos \alpha}{1 - \sin \alpha}\right) \]

### Step 5: Express in the required form
Now we can express the complex number in the desired form:
\[ 1 - \sin \alpha + i \cos \alpha = r(\cos \theta + i \sin \theta) \]
Substituting the values of \(r\) and \(\theta\):
\[ 1 - \sin \alpha + i \cos \alpha = \sqrt{2(1 - \sin \alpha)}\left(\cos\left(\tan^{-1}\left(\frac{\cos \alpha}{1 - \sin \alpha}\right)\right) + i \sin\left(\tan^{-1}\left(\frac{\cos \alpha}{1 - \sin \alpha}\right)\right)\right) \]

### Final Result
Thus, the expression of the complex number \(1 - \sin \alpha + i \cos \alpha\) in the form \(r(\cos \theta + i \sin \theta)\) is:
\[ 1 - \sin \alpha + i \cos \alpha = \sqrt{2(1 - \sin \alpha)}\left(\cos\left(\tan^{-1}\left(\frac{\cos \alpha}{1 - \sin \alpha}\right)\right) + i \sin\left(\tan^{-1}\left(\frac{\cos \alpha}{1 - \sin \alpha}\right)\right)\right) \]