`cot^(-1)((1+cosx)/(sinx))`

Answer

To solve the problem of differentiating \( y = \cot^{-1}\left(\frac{1 + \cos x}{\sin x}\right) \), we will follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ y = \cot^{-1}\left(\frac{1 + \cos x}{\sin x}\right) \]

### Step 2: Simplify the Argument
Using the trigonometric identities, we can simplify the expression inside the cotangent inverse. We know that:
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
and
\[ \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]
Thus, we can rewrite the argument of the cotangent inverse:
\[ \frac{1 + \cos x}{\sin x} = \frac{2 \cos^2\left(\frac{x}{2}\right)}{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)} = \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} = \cot\left(\frac{x}{2}\right) \]

### Step 3: Substitute Back into the Function
Now we can substitute this back into our expression for \( y \):
\[ y = \cot^{-1}\left(\cot\left(\frac{x}{2}\right)\right) \]
Since \( \cot^{-1}(\cot(\theta)) = \theta \) for \( \theta \) in the appropriate range, we have:
\[ y = \frac{x}{2} \]

### Step 4: Differentiate with Respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \]

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