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

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

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

### Statement 1:
If \( y = \ln(\sec x + \tan x) \), then \( \frac{dy}{dx} = \sec x \).

**Step 1:** Differentiate \( y \) using the chain rule.
\[ \frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x) \]

**Step 2:** Differentiate \( \sec x \) and \( \tan x \).
- The derivative of \( \sec x \) is \( \sec x \tan x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).

Thus,
\[ \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x \]

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

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

**Step 5:** Simplify the expression.
Since \( \tan x + \sec x = \sec x + \tan x \), we have:
\[ \frac{dy}{dx} = \sec x \]

Thus, **Statement 1 is correct**.

### Statement 2:
If \( y = \ln(\csc x - \cot x) \), then \( \frac{dy}{dx} = \csc x \).

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

**Step 2:** Differentiate \( \csc x \) and \( \cot x \).
- The derivative of \( \csc x \) is \( -\csc x \cot x \).
- The derivative of \( \cot x \) is \( -\csc^2 x \).

Thus,
\[ \frac{d}{dx}(\csc x - \cot x) = -\csc x \cot x + \csc^2 x \]

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

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

**Step 5:** Simplify the expression.
Since \( -\cot x + \csc x = \csc x - \cot x \), we have:
\[ \frac{dy}{dx} = \csc x \]

Thus, **Statement 2 is also correct**.

### Conclusion:
Both statements are correct.