As $\theta$ varies, what's the largest this ratio of areas can be?

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Let \(a > 0\). Sketch the graph of

\[y=\dfrac{a+x}{a-x} \quad \text{for} \quad -a < x < a.\]

Let \(0 < \theta < \dfrac{\pi}{2}\). In the diagram below is the half-disc given by \(x^2 + y^2 \leq 1\) and \(y \geq 0\).

The shaded region \(A\) consists of those points with \(-\cos \theta \leq x \leq \sin \theta\). The region \(B\) is the remainder of the half-disc.

Find the area of \(A\).

Assuming only that \(\sin^2\theta + \cos^2\theta = 1\), show that \(\sin\theta\cos\theta \leq \dfrac{1}{2}\).

What is the largest \(\dfrac{\text{area of } A}{\text{area of } B}\) can be, as \(\theta\) varies?