Determine value of real parameter $\lambda$ such that equation $$\frac{1}{\sin{x}} + \frac{1}{\cos{x}} = \lambda $$ has root in interval $\left(0,\frac{\pi}{2}\right)$
1. **Identify the domain and behavior of the trigonometric functions:**
   Since we are considering the interval \( \left(0, \frac{\pi}{2}\right) \), both \(\sin x\) and \(\cos x\) are positive in this interval. Therefore, \(\frac{1}{\sin x}\) and \(\frac{1}{\cos x}\) are also positive.

2. **Apply the Power Mean Inequality:**
   The Power Mean Inequality states that for non-negative real numbers \(a\) and \(b\),
   \[
   \frac{a + b}{2} \geq \sqrt{ab}
   \]
   with equality if and only if \(a = b\). Applying this to \(\frac{1}{\sin x}\) and \(\frac{1}{\cos x}\), we get:
   \[
   \frac{\frac{1}{\sin x} + \frac{1}{\cos x}}{2} \geq \sqrt{\frac{1}{\sin x} \cdot \frac{1}{\cos x}}
   \]
   Simplifying the right-hand side:
   \[
   \sqrt{\frac{1}{\sin x \cos x}} = \sqrt{\frac{2}{\sin 2x}}
   \]
   Since \(\sin 2x \leq 1\) for all \(x\), we have:
   \[
   \sqrt{\frac{2}{\sin 2x}} \geq \sqrt{2}
   \]
   Therefore:
   \[
   \frac{\frac{1}{\sin x} + \frac{1}{\cos x}}{2} \geq \sqrt{2}
   \]
   Multiplying both sides by 2:
   \[
   \frac{1}{\sin x} + \frac{1}{\cos x} \geq 2\sqrt{2}
   \]
   Equality occurs when \(\sin x = \cos x\), which happens at \(x = \frac{\pi}{4}\).

3. **Continuity and Intermediate Value Theorem (IVT):**
   The function \(\frac{1}{\sin x} + \frac{1}{\cos x}\) is continuous on \(\left(0, \frac{\pi}{2}\right)\). As \(x \to 0^+\), \(\sin x \to 0\) and \(\cos x \to 1\), so \(\frac{1}{\sin x} \to \infty\) and \(\frac{1}{\cos x} \to 1\). Thus:
   \[
   \lim_{x \to 0^+} \left( \frac{1}{\sin x} + \frac{1}{\cos x} \right) = +\infty
   \]
   At \(x = \frac{\pi}{4}\), we have:
   \[
   \frac{1}{\sin \frac{\pi}{4}} + \frac{1}{\cos \frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} + \frac{1}{\frac{\sqrt{2}}{2}} = 2\sqrt{2}
   \]
   By the Intermediate Value Theorem, since \(\frac{1}{\sin x} + \frac{1}{\cos x}\) is continuous and takes on all values from \(2\sqrt{2}\) to \(+\infty\) as \(x\) moves from \(\frac{\pi}{4}\) to \(0\), there exists some \(x \in \left(0, \frac{\pi}{4}\right)\) such that \(\frac{1}{\sin x} + \frac{1}{\cos x} = \lambda\) for any \(\lambda \geq 2\sqrt{2}\).

Conclusion:
The equation \(\frac{1}{\sin x} + \frac{1}{\cos x} = \lambda\) has a root in the interval \(\left(0, \frac{\pi}{2}\right)\) if and only if \(\lambda \geq 2\sqrt{2}\).

The final answer is \( \boxed{ \lambda \geq 2\sqrt{2} } \)