If `y=log sqrt((1-cos x)/(1+ cos x))`, then `(dy)/(dx)` is

Options

A. `-sin x`
B. `- "cosec"x`
C. sin x
D. cosec x

Answer

To find the derivative of the function \( y = \log \sqrt{\frac{1 - \cos x}{1 + \cos x}} \), we will follow these steps:

### Step 1: Simplify the expression
We can rewrite the logarithm of the square root as:
\[ y = \frac{1}{2} \log \left( \frac{1 - \cos x}{1 + \cos x} \right) \]

### Step 2: Use the properties of logarithms
Using the properties of logarithms, we can separate the fraction:
\[ y = \frac{1}{2} \left( \log(1 - \cos x) - \log(1 + \cos x) \right) \]

### Step 3: Differentiate using the chain rule
Now, we will differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{d}{dx} \log(1 - \cos x) - \frac{d}{dx} \log(1 + \cos x) \right) \]

Using the derivative of the logarithm \( \frac{d}{dx} \log(u) = \frac{1}{u} \cdot \frac{du}{dx} \):
- For \( \log(1 - \cos x) \):
\[ \frac{d}{dx} \log(1 - \cos x) = \frac{1}{1 - \cos x} \cdot \sin x \]
- For \( \log(1 + \cos x) \):
\[ \frac{d}{dx} \log(1 + \cos x) = \frac{1}{1 + \cos x} \cdot (-\sin x) \]

### Step 4: Substitute back into the derivative
Now substituting these derivatives back into our expression:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{\sin x}{1 - \cos x} + \frac{\sin x}{1 + \cos x} \right) \]

### Step 5: Combine the fractions
Combining the two fractions:
\[ \frac{dy}{dx} = \frac{1}{2} \sin x \left( \frac{1}{1 - \cos x} + \frac{1}{1 + \cos x} \right) \]
Finding a common denominator:
\[ \frac{dy}{dx} = \frac{1}{2} \sin x \left( \frac{(1 + \cos x) + (1 - \cos x)}{(1 - \cos x)(1 + \cos x)} \right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{2} \sin x \left( \frac{2}{1 - \cos^2 x} \right) \]

### Step 6: Use the identity \( 1 - \cos^2 x = \sin^2 x \)
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{2} \sin x \left( \frac{2}{\sin^2 x} \right) = \frac{1}{\sin x} \]

### Final Result
Therefore, the derivative is:
\[ \frac{dy}{dx} = \csc x \]