Statement-1: `f(x ) = log_( cosx)sinx` is well defined in `(0,pi/2)`.
and Statement:2 sinx and cosx are positive in `(0,pi/2)`

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-2
B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct expanation for statement-2
C. Statement-1 is True, Statement-2 is false
D. Statement-1 is False, Statement-2 is true

Answer

To determine whether the statements are true, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
The function given is \( f(x) = \log_{\cos x}(\sin x) \). We need to check if this function is well-defined in the interval \( (0, \frac{\pi}{2}) \).

#### Conditions for the logarithm:
1. The argument of the logarithm must be positive: \( \sin x > 0 \).
2. The base of the logarithm must be positive and not equal to 1: \( \cos x > 0 \) and \( \cos x \neq 1 \).

### Step 2: Check the conditions in the interval \( (0, \frac{\pi}{2}) \)
- **For \( \sin x > 0 \)**:
- In the interval \( (0, \frac{\pi}{2}) \), \( \sin x \) is positive since \( \sin x \) increases from 0 to 1.

- **For \( \cos x > 0 \)**:
- In the interval \( (0, \frac{\pi}{2}) \), \( \cos x \) is also positive since \( \cos x \) decreases from 1 to 0.

- **For \( \cos x \neq 1 \)**:
- At \( x = 0 \), \( \cos(0) = 1 \), but since we are considering the open interval \( (0, \frac{\pi}{2}) \), this condition is satisfied.

### Conclusion for Statement 1:
Since both conditions are satisfied in the interval \( (0, \frac{\pi}{2}) \), **Statement 1 is true**.

### Step 3: Analyze Statement 2
Statement 2 claims that \( \sin x \) and \( \cos x \) are positive in the interval \( (0, \frac{\pi}{2}) \).

- **For \( \sin x \)**:
- As discussed, \( \sin x \) is positive in the interval \( (0, \frac{\pi}{2}) \).

- **For \( \cos x \)**:
- Similarly, \( \cos x \) is also positive in the interval \( (0, \frac{\pi}{2}) \).

### Conclusion for Statement 2:
Since both \( \sin x \) and \( \cos x \) are positive in the interval \( (0, \frac{\pi}{2}) \), **Statement 2 is also true**.

### Final Conclusion:
Both statements are true.

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