If `cos x ne -1`. then what is `(sinx)/(1+cosx)` equal to?

Options

A. `-cot[(x)/(2)]`
B. `cot[(x)/(2)]`
C. `tan[(x)/(2)]`
D. `-tan [(x)/(2)]`

Answer

To solve the problem of finding the value of \(\frac{\sin x}{1 + \cos x}\) given that \(\cos x \neq -1\), we can follow these steps:

### Step 1: Use the double angle identity for sine
We know the double angle identity for sine:
\[ \sin 2x = 2 \sin x \cos x \]
However, we can also express \(\sin x\) in terms of half-angle identities:
\[ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \]

### Step 2: Rewrite \(1 + \cos x\) using the half-angle identity
Using the half-angle identity for cosine, we have:
\[ \cos x = 2 \cos^2 \frac{x}{2} - 1 \]
Thus,
\[ 1 + \cos x = 1 + (2 \cos^2 \frac{x}{2} - 1) = 2 \cos^2 \frac{x}{2} \]

### Step 3: Substitute into the original expression
Now we can substitute the expressions for \(\sin x\) and \(1 + \cos x\) into our original expression:
\[ \frac{\sin x}{1 + \cos x} = \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \]

### Step 4: Simplify the expression
The \(2\) in the numerator and denominator cancels out:
\[ \frac{\sin \frac{x}{2} \cos \frac{x}{2}}{\cos^2 \frac{x}{2}} \]
Now, we can simplify further:
\[ = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \]

### Step 5: Recognize the result
The expression \(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\) is the definition of the tangent function:
\[ = \tan \frac{x}{2} \]

### Final Answer
Thus, we conclude that:
\[ \frac{\sin x}{1 + \cos x} = \tan \frac{x}{2} \]