1. The problem is to plot the function $$f(x) = \left(\frac{1 + \sin x}{1 - \sin x}\right)^{\frac{1}{2}}$$ on the y-axis. 2. This function involves a square root of a fraction where the numerator is $1 + \sin x$ and the denominator is $1 - \sin x$. 3. Important to note: The denominator $1 - \sin x$ must not be zero, and the expression inside the square root must be non-negative for the function to be real. 4. Simplify the function if possible. Using the identity for tangent of half-angle, we can rewrite: $$\frac{1 + \sin x}{1 - \sin x} = \left(\frac{1 + \sin x}{1 - \sin x}\right) = \left(\frac{1 + \sin x}{1 - \sin x}\right)$$ 5. Recognize that $$\frac{1 + \sin x}{1 - \sin x} = \left(\frac{\cos x + \sin x}{\cos x - \sin x}\right)^2$$ but to keep it simple, we will plot the original function. 6. The function is defined where $1 - \sin x \neq 0$ and $\frac{1 + \sin x}{1 - \sin x} \geq 0$. 7. The function can be plotted as $$y = \sqrt{\frac{1 + \sin x}{1 - \sin x}}$$. Final answer: $$f(x) = \sqrt{\frac{1 + \sin x}{1 - \sin x}}$$