If `y=log((1-cos x)/(1+cos x))` thenn `dy/dx=`

Options

A. ` -cosec x `
B. ` cosecx`
C. `-2 cosecx`
D. `2 cosecx`

Answer

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \log\left(\frac{1 - \cos x}{1 + \cos x}\right) \), we will use the properties of logarithms and differentiation.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = \log\left(\frac{1 - \cos x}{1 + \cos x}\right) \]

2. **Use the logarithmic differentiation**:
The derivative of \( \log(f(x)) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{f(x)} \cdot f'(x) \]
where \( f(x) = \frac{1 - \cos x}{1 + \cos x} \).

3. **Differentiate \( f(x) \)**:
To differentiate \( f(x) \), we will use the quotient rule:
\[ f'(x) = \frac{(g(x) \cdot f'(x)) - (f(x) \cdot g'(x))}{(g(x))^2} \]
where \( f(x) = 1 - \cos x \) and \( g(x) = 1 + \cos x \).

- Here, \( f'(x) = \sin x \) and \( g'(x) = -\sin x \).

Thus, we have:
\[ f'(x) = \frac{(1 + \cos x)(\sin x) - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2} \]

4. **Simplify the expression**:
\[ f'(x) = \frac{(1 + \cos x)\sin x + (1 - \cos x)\sin x}{(1 + \cos x)^2} \]
\[ = \frac{\sin x (1 + \cos x + 1 - \cos x)}{(1 + \cos x)^2} \]
\[ = \frac{2\sin x}{(1 + \cos x)^2} \]

5. **Substitute back into the derivative of \( y \)**:
Now substituting \( f(x) \) and \( f'(x) \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{\frac{1 - \cos x}{1 + \cos x}} \cdot \frac{2\sin x}{(1 + \cos x)^2} \]
\[ = \frac{(1 + \cos x)}{(1 - \cos x)} \cdot \frac{2\sin x}{(1 + \cos x)^2} \]
\[ = \frac{2\sin x}{(1 - \cos x)(1 + \cos x)} \]

6. **Simplify further**:
Using the identity \( 1 - \cos^2 x = \sin^2 x \):
\[ = \frac{2\sin x}{\sin^2 x} = \frac{2}{\sin x} \]

7. **Final result**:
\[ \frac{dy}{dx} = 2 \csc x \]