Let`f'(sin x)lt0 and f''(sin x) gt0 forall x in (0,(pi)/(2))` and g(x) =f(sinx)+f(cosx)
which of the following is true?

Options

A. g(x) is decreasing in `((pi)/(4),(pi)/(2))`
B. g(x) increasing in `(0,(pi)/(4))`
C. g(x) is nonotonically increasing in `(0,(pi)/(2))`
D. none of these

Explanation

To solve the problem, we will analyze the function \( g(x) = f(\sin x) + f(\cos x) \) using the given conditions about the derivatives of \( f \).

### Step 1: Understand the given conditions
We have:
1. \( f'(\sin x) < 0 \) for all \( x \in (0, \frac{\pi}{2}) \)
2. \( f''(\sin x) > 0 \) for all \( x \in (0, \frac{\pi}{2}) \)

From these conditions, we can infer that:
- Since \( f'(\sin x) < 0 \), the function \( f \) is decreasing on the interval \( (0, 1) \) (as \( \sin x \) ranges from 0 to 1).
- Since \( f''(\sin x) > 0 \), the function \( f' \) is increasing, which means that \( f \) is concave up on this interval.

### Step 2: Analyze \( g(x) \)
We need to find the behavior of the function \( g(x) = f(\sin x) + f(\cos x) \).

### Step 3: Differentiate \( g(x) \)
To analyze the monotonicity of \( g(x) \), we will differentiate it:
\[ g'(x) = f'(\sin x) \cos x - f'(\cos x) \sin x \]
This uses the chain rule for differentiation.

### Step 4: Analyze the sign of \( g'(x) \)
Given that \( f'(\sin x) < 0 \) and \( f'(\cos x) < 0 \) in the interval \( (0, \frac{\pi}{2}) \), we need to analyze the expression \( g'(x) \):
\[ g'(x) = f'(\sin x) \cos x - f'(\cos x) \sin x \]
- Since \( f'(\sin x) < 0 \) and \( \cos x > 0 \) in \( (0, \frac{\pi}{2}) \), the term \( f'(\sin x) \cos x < 0 \).
- Similarly, since \( f'(\cos x) < 0 \) and \( \sin x > 0 \) in \( (0, \frac{\pi}{2}) \), the term \( -f'(\cos x) \sin x < 0 \).

### Step 5: Determine the behavior of \( g(x) \)
Now we need to check the intervals:
1. For \( x \in (0, \frac{\pi}{4}) \):
- Both terms are negative, but we need to determine which term dominates.
- As \( x \) approaches \( \frac{\pi}{4} \), \( \sin x \) and \( \cos x \) are equal, and since \( f' \) is decreasing, we can conclude that \( g'(x) < 0 \) in this interval, meaning \( g(x) \) is decreasing.

2. For \( x \in (\frac{\pi}{4}, \frac{\pi}{2}) \):
- As \( x \) approaches \( \frac{\pi}{2} \), \( \sin x \) approaches 1 and \( \cos x \) approaches 0.
- Here, \( g'(x) \) will also be negative, meaning \( g(x) \) continues to decrease.

### Conclusion
Thus, we can conclude:
- \( g(x) \) is decreasing in the interval \( (0, \frac{\pi}{2}) \).
- Specifically, \( g(x) \) is decreasing in \( (0, \frac{\pi}{4}) \) and continues to decrease in \( (\frac{\pi}{4}, \frac{\pi}{2}) \).

### Final Answer
None of the provided options are true, as \( g(x) \) is decreasing throughout the interval \( (0, \frac{\pi}{2}) \).