Consider the following statements
I. If `y=ln(secx+tanx)`. then `(dy)/(dx)=secx`
II. If `y=ln("cosec"x-cotx)`. then `(dy)/(dx)="cosec"x`
Which of the above statements is/are correct ?

Options

A. Only I
B. Only 2
C. Both 1 and 2
D. Neither 1 nor 2

Answer

To solve the problem, we need to verify the correctness of the two statements regarding derivatives of logarithmic functions. Let's go through each statement step by step.

### Statement I:
**If \( y = \ln(\sec x + \tan x) \), then \( \frac{dy}{dx} = \sec x \)**

1. **Differentiate \( y \)**:
\[ y = \ln(\sec x + \tan x) \]
Using the chain rule, the derivative is:
\[ \frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x) \]

2. **Differentiate \( \sec x + \tan x \)**:
The derivatives of \( \sec x \) and \( \tan x \) are:
\[ \frac{d}{dx}(\sec x) = \sec x \tan x \]
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]
Therefore:
\[ \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x \]

3. **Substituting back**:
\[ \frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) \]

4. **Factor out \( \sec x \)**:
\[ \frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \]

5. **Simplify**:
The numerator and denominator are the same:
\[ \frac{dy}{dx} = \sec x \]

Thus, **Statement I is correct**.

### Statement II:
**If \( y = \ln(\csc x - \cot x) \), then \( \frac{dy}{dx} = \csc x \)**

1. **Differentiate \( y \)**:
\[ y = \ln(\csc x - \cot x) \]
Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{\csc x - \cot x} \cdot \frac{d}{dx}(\csc x - \cot x) \]

2. **Differentiate \( \csc x - \cot x \)**:
The derivatives of \( \csc x \) and \( \cot x \) are:
\[ \frac{d}{dx}(\csc x) = -\csc x \cot x \]
\[ \frac{d}{dx}(\cot x) = -\csc^2 x \]
Therefore:
\[ \frac{d}{dx}(\csc x - \cot x) = -\csc x \cot x + \csc^2 x \]

3. **Substituting back**:
\[ \frac{dy}{dx} = \frac{1}{\csc x - \cot x} \cdot (-\csc x \cot x + \csc^2 x) \]

4. **Factor out \( \csc x \)**:
\[ \frac{dy}{dx} = \frac{\csc x (\csc x - \cot x)}{\csc x - \cot x} \]

5. **Simplify**:
The numerator and denominator are the same:
\[ \frac{dy}{dx} = \csc x \]

Thus, **Statement II is also correct**.

### Conclusion:
Both statements are correct. Therefore, the answer is that both statements are true.