Let ` f(x)= sinx - tanx, x in (0, pi//2) ` then tangent drawn to the curve ` y= f(x) ` at any point will

Options

A. lie above the curve
B. lie below the curve
C. nothing can be said
D. be parallel to a fixed line.

Explanation

To solve the problem, we need to analyze the function \( f(x) = \sin x - \tan x \) for \( x \) in the interval \( (0, \frac{\pi}{2}) \) and determine the position of the tangent line drawn at any point on the curve \( y = f(x) \).

### Step 1: Differentiate the function
First, we find the first derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin x - \tan x) = \cos x - \sec^2 x \]

### Step 2: Find the second derivative
Next, we differentiate \( f'(x) \) to find the second derivative:
\[ f''(x) = \frac{d}{dx}(\cos x - \sec^2 x) = -\sin x - 2\sec^2 x \tan x \]

### Step 3: Analyze the second derivative
We need to analyze \( f''(x) \) in the interval \( (0, \frac{\pi}{2}) \):
- Since \( \sin x \) is positive and \( \sec^2 x \) is also positive for \( x \) in \( (0, \frac{\pi}{2}) \), it follows that \( f''(x) \) is negative.
- Therefore, \( f''(x) < 0 \) implies that the function \( f(x) \) is concave down in this interval.

### Step 4: Conclusion about the tangent line
Since \( f(x) \) is concave down, the tangent line at any point on the curve will lie above the curve itself. This is because, for a concave down function, the tangent line at any point will not intersect the curve again in the interval.

### Final Result
Thus, we conclude that the tangent drawn to the curve \( y = f(x) \) at any point will be above the curve.