1. **State the problem:** We want to study the function $$f(x) = \sqrt{\frac{1+\sin x}{1-\sin x}}$$. 2. **Recall important formulas and rules:** - The function involves a square root, so the expression inside must be non-negative: $$\frac{1+\sin x}{1-\sin x} \geq 0$$. - The denominator cannot be zero: $$1 - \sin x \neq 0$$. - The sine function ranges between -1 and 1. 3. **Analyze the domain:** - Denominator zero when $$1 - \sin x = 0 \Rightarrow \sin x = 1$$, which happens at $$x = \frac{\pi}{2} + 2k\pi$$, $k \in \mathbb{Z}$$, so these points are excluded. - For the fraction to be non-negative: - Both numerator and denominator positive, or - Both numerator and denominator negative. 4. **Case 1: numerator and denominator both positive:** - $$1 + \sin x > 0 \Rightarrow \sin x > -1$$ (always true since $$\sin x \geq -1$$) - $$1 - \sin x > 0 \Rightarrow \sin x < 1$$ - So for $$\sin x < 1$$ and $$\sin x > -1$$, fraction positive except at $$\sin x = 1$$. 5. **Case 2: numerator and denominator both negative:** - $$1 + \sin x < 0 \Rightarrow \sin x < -1$$ (impossible) - So no solution here. 6. **Domain summary:** - $$\sin x \neq 1$$ - Since $$\sin x$$ is always between -1 and 1, the domain is all real $$x$$ except where $$\sin x = 1$$. 7. **Simplify the function:** Use the identity $$\tan\left(\frac{x}{2} + \frac{\pi}{4}\right) = \sqrt{\frac{1+\sin x}{1-\sin x}}$$, so $$f(x) = \left|\tan\left(\frac{x}{2} + \frac{\pi}{4}\right)\right|$$. 8. **Behavior and range:** - Since $$f(x)$$ is absolute value of tangent shifted, it takes all positive real values. 9. **Final answer:** The function $$f(x) = \sqrt{\frac{1+\sin x}{1-\sin x}}$$ is defined for all real $$x$$ except where $$\sin x = 1$$ (i.e., $$x = \frac{\pi}{2} + 2k\pi$$). It can be rewritten as $$f(x) = \left|\tan\left(\frac{x}{2} + \frac{\pi}{4}\right)\right|$$. This helps understand its behavior and range.