The given expression is \(x=\frac{\sin ^2 A+\sin A+1}{\sin A}\), and the range of angle A is \(0 < A ≤ \frac{\pi}{2}\).

Simplifying the Trigonometric Expression

We can simplify the given expression for x by dividing each term in the numerator by \(\sin A\):

\[x = \frac{\sin^2 A}{\sin A} + \frac{\sin A}{\sin A} + \frac{1}{\sin A}\]

\[x = \sin A + 1 + \frac{1}{\sin A}\]

Understanding the Range of sin A

The given range for A is \(0 < A ≤ \frac{\pi}{2}\). In this range, the value of \(\sin A\) is positive and increases from a value close to 0 (but not zero) up to 1. Specifically, for \(0 < A ≤ \frac{\pi}{2}\), the range of \(\sin A\) is \(0 < \sin A ≤ 1\).

Let \(y = \sin A\). Our expression becomes \(x = y + 1 + \frac{1}{y}\), and we need to find the value of A (which corresponds to a value of y in \(0 < y ≤ 1\)) that minimizes this expression.

Applying AM-GM Inequality for Minimum Value

To find the minimum value of \(y + \frac{1}{y}\) for \(y > 0\), we can use the Arithmetic Mean - Geometric Mean (AM-GM) inequality. The AM-GM inequality states that for any two non-negative numbers a and b, the arithmetic mean is greater than or equal to the geometric mean: \(\frac{a+b}{2} \ge \sqrt{ab}\). Equality holds if and only if \(a=b\).

Applying this to \(y\) and \(\frac{1}{y}\) (both are positive since \(0 < y \le 1\)):

\[y + \frac{1}{y} \ge 2\sqrt{y \cdot \frac{1}{y}}\]

\[y + \frac{1}{y} \ge 2\sqrt{1}\]

\[y + \frac{1}{y} \ge 2\]

The minimum value of \(y + \frac{1}{y}\) for \(y > 0\) is 2.

This minimum is achieved when \(y = \frac{1}{y}\), which means \(y^2 = 1\). Since \(y = \sin A\) and \(0 < A ≤ \frac{\pi}{2}\), \(y\) must be positive. Thus, the minimum occurs when \(y=1\).

Finding the Value of A for Minimum x

The expression for x is \(x = y + \frac{1}{y} + 1\). The minimum value of \(y + \frac{1}{y}\) is 2, occurring when \(y=1\). Therefore, the minimum value of x is \(2 + 1 = 3\), which occurs when \(y = 1\).

We defined \(y = \sin A\). So, the minimum value of x occurs when \(\sin A = 1\).

We need to find the value of A in the given range \(0 < A ≤ \frac{\pi}{2}\) for which \(\sin A = 1\). The angle in this range whose sine is 1 is \(A = \frac{\pi}{2}\).

Conclusion: Minimum Value of x

The minimum value of the expression \(x\) is attained when \(A = \frac{\pi}{2}\).

Revision Table: Finding Minimum of Trigonometric Functions

Additional Information: Properties of sin A and AM-GM

In the range \(0 < A ≤ \frac{\pi}{2}\), the sine function starts from values close to 0 and increases monotonically to 1 at \(A = \frac{\pi}{2}\). This behaviour is crucial for determining the range of \(y = \sin A\).

The AM-GM inequality is a powerful tool for finding the minimum or maximum of expressions involving sums and products of positive numbers. It states that the arithmetic mean is always greater than or equal to the geometric mean, with equality holding only when all terms are equal. In this problem, applying it to \(y\) and \(1/y\) directly gives the minimum of their sum.

Another approach could involve calculus by considering \(f(y) = y + \frac{1}{y} + 1\) and finding its derivative with respect to \(y\), \(f'(y) = 1 - \frac{1}{y^2}\). Setting \(f'(y) = 0\) gives \(y^2 = 1\). Since \(y > 0\), \(y=1\). Evaluating the function at the boundary \(y=1\) and considering the limit as \(y \to 0^+\) would also lead to the conclusion that the minimum on the interval \(0 < y \le 1\) occurs at \(y=1\).