Simplify
`tan^(-1)((sinx)/(1+cos x))`

Answer

To simplify the expression \( \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) \), we can follow these steps:

### Step 1: Rewrite sine and cosine in terms of tangent
We know the following identities for sine and cosine in terms of tangent:
\[ \sin x = \frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
\[ \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]

### Step 2: Substitute sine and cosine into the expression
Now, substituting these identities into the expression:
\[ \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) = \tan^{-1}\left(\frac{\frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}}{1 + \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}}\right) \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:
\[ 1 + \cos x = 1 + \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} = \frac{(1 + \tan^2\left(\frac{x}{2}\right)) + (1 - \tan^2\left(\frac{x}{2}\right))}{1 + \tan^2\left(\frac{x}{2}\right)} = \frac{2}{1 + \tan^2\left(\frac{x}{2}\right)} \]

### Step 4: Substitute back into the expression
Now substituting this back into our expression:
\[ \tan^{-1}\left(\frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \cdot \frac{1 + \tan^2\left(\frac{x}{2}\right)}{2}\right) \]

### Step 5: Cancel out terms
The \(1 + \tan^2\left(\frac{x}{2}\right)\) cancels out:
\[ \tan^{-1}\left(\frac{2 \tan\left(\frac{x}{2}\right)}{2}\right) = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) \]

### Step 6: Apply the inverse tangent identity
Using the identity \( \tan^{-1}(\tan \theta) = \theta \):
\[ \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) = \frac{x}{2} \]

### Final Answer
Thus, the simplified expression is:
\[ \frac{x}{2} \]

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