If `0ltxlt(pi)/(2)` then the minimum value of `(cos^(3)x)/(sinx)+(sin^(3))/(cosx)(x)/(x)`, is

Options

A. `sqrt(3)`
B. `(1)/(2)`
C. `(1)/(3)`
D. 1

Answer

To find the minimum value of the expression \(\frac{\cos^3 x}{\sin x} + \frac{\sin^3 x}{\cos x}\) for \(0 < x < \frac{\pi}{2}\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Identify the terms**: We have two terms in our expression:
\[ a = \frac{\cos^3 x}{\sin x}, \quad b = \frac{\sin^3 x}{\cos x} \]

2. **Apply AM-GM Inequality**: According to the AM-GM inequality:
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
Therefore, we can write:
\[ \frac{\frac{\cos^3 x}{\sin x} + \frac{\sin^3 x}{\cos x}}{2} \geq \sqrt{\frac{\cos^3 x}{\sin x} \cdot \frac{\sin^3 x}{\cos x}} \]

3. **Simplify the right side**:
\[ ab = \frac{\cos^3 x \cdot \sin^3 x}{\sin x \cdot \cos x} = \cos^2 x \sin^2 x \]
Thus,
\[ \sqrt{ab} = \sqrt{\cos^2 x \sin^2 x} = \cos x \sin x \]

4. **Substitute back into the AM-GM inequality**:
\[ \frac{\frac{\cos^3 x}{\sin x} + \frac{\sin^3 x}{\cos x}}{2} \geq \cos x \sin x \]
This implies:
\[ \frac{\cos^3 x}{\sin x} + \frac{\sin^3 x}{\cos x} \geq 2 \cos x \sin x \]

5. **Express \(2 \cos x \sin x\)**: We know that:
\[ 2 \cos x \sin x = \sin(2x) \]
Therefore:
\[ \frac{\cos^3 x}{\sin x} + \frac{\sin^3 x}{\cos x} \geq \sin(2x) \]

6. **Find the maximum of \(\sin(2x)\)**: The maximum value of \(\sin(2x)\) occurs at \(x = \frac{\pi}{4}\) where:
\[ \sin(2x) = \sin\left(\frac{\pi}{2}\right) = 1 \]

7. **Conclusion**: Thus, we have:
\[ \frac{\cos^3 x}{\sin x} + \frac{\sin^3 x}{\cos x} \geq 1 \]
The minimum value of the expression \(\frac{\cos^3 x}{\sin x} + \frac{\sin^3 x}{\cos x}\) is therefore:
\[ \boxed{1} \]