If inequality $ \frac {\sin ^{3} x}{\cos x} \plus{} \frac {\cos ^{3} x}{\sin x} \ge k$ is hold for every $ x\in \left(0,\frac {\pi }{2} \right)$, what is the largest possible value of $ k$?

$\textbf{(A)}\  \frac {1}{2} \qquad\textbf{(B)}\  \frac {3}{4} \qquad\textbf{(C)}\ 1 \qquad\textbf{(D)}\  \frac {3}{2} \qquad\textbf{(E)}\ \text{None}$
To solve the given inequality \( \frac{\sin^3 x}{\cos x} + \frac{\cos^3 x}{\sin x} \ge k \) for \( x \in \left(0, \frac{\pi}{2}\right) \), we will use the AM-GM inequality and some trigonometric identities.

1. **Rewrite the expression:**
   \[
   \frac{\sin^3 x}{\cos x} + \frac{\cos^3 x}{\sin x} = \sin^2 x \cdot \frac{\sin x}{\cos x} + \cos^2 x \cdot \frac{\cos x}{\sin x} = \sin^2 x \tan x + \cos^2 x \cot x
   \]

2. **Apply the AM-GM inequality:**
   The AM-GM inequality states that for any non-negative real numbers \(a\) and \(b\),
   \[
   \frac{a + b}{2} \ge \sqrt{ab}
   \]
   Applying this to our terms, we get:
   \[
   \frac{\sin^2 x \tan x + \cos^2 x \cot x}{2} \ge \sqrt{\sin^2 x \tan x \cdot \cos^2 x \cot x}
   \]

3. **Simplify the right-hand side:**
   \[
   \sqrt{\sin^2 x \tan x \cdot \cos^2 x \cot x} = \sqrt{\sin^2 x \cdot \frac{\sin x}{\cos x} \cdot \cos^2 x \cdot \frac{\cos x}{\sin x}} = \sqrt{\sin^2 x \cos^2 x}
   \]
   Since \(\sin x \cos x\) is always positive in \((0, \frac{\pi}{2})\), we have:
   \[
   \sqrt{\sin^2 x \cos^2 x} = \sin x \cos x
   \]

4. **Combine the results:**
   \[
   \frac{\sin^2 x \tan x + \cos^2 x \cot x}{2} \ge \sin x \cos x
   \]
   Therefore,
   \[
   \sin^2 x \tan x + \cos^2 x \cot x \ge 2 \sin x \cos x
   \]

5. **Find the minimum value:**
   We need to find the largest possible value of \(k\) such that:
   \[
   \sin^2 x \tan x + \cos^2 x \cot x \ge k
   \]
   From the previous step, we have:
   \[
   \sin^2 x \tan x + \cos^2 x \cot x \ge 2 \sin x \cos x
   \]
   The minimum value of \(2 \sin x \cos x\) occurs when \(\sin x = \cos x = \frac{\sqrt{2}}{2}\), giving:
   \[
   2 \sin x \cos x = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1
   \]
   Therefore, the largest possible value of \(k\) is \(1\).

The final answer is \(\boxed{1}\)