If `y = tan^(-1) ((1 - cos x)/(sin x)) " then "(dy)/(dx)=` ?

Options

A. 1
B. `-1`
C. `(1)/(2)`
D. `(-1)/(2)`

Explanation

To solve the problem, we need to differentiate the function \( y = \tan^{-1} \left( \frac{1 - \cos x}{\sin x} \right) \).

### Step-by-Step Solution:

1. **Rewrite the Function**:
We start with the function:
\[ y = \tan^{-1} \left( \frac{1 - \cos x}{\sin x} \right) \]

2. **Use Trigonometric Identities**:
We can use the identity \( 1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right) \) to rewrite \( 1 - \cos x \):
\[ y = \tan^{-1} \left( \frac{2 \sin^2 \left( \frac{x}{2} \right)}{\sin x} \right) \]
We also know that \( \sin x = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \). Thus, we can rewrite \( y \) as:
\[ y = \tan^{-1} \left( \frac{2 \sin^2 \left( \frac{x}{2} \right)}{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} \right) \]

3. **Simplify the Expression**:
The \( 2 \) in the numerator and denominator cancels out:
\[ y = \tan^{-1} \left( \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)} \right) \]
This simplifies to:
\[ y = \tan^{-1} \left( \tan \left( \frac{x}{2} \right) \right) \]
Since \( \tan^{-1} \tan \theta = \theta \) for \( \theta \) in the principal range, we have:
\[ y = \frac{x}{2} \]

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

### Final Answer:
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \]