1. **Stating the problem:** We want to analyze the function $$y = \ln\left(\frac{\sin x - 1}{\sin x + 1}\right)$$ and understand its domain and behavior. 2. **Formula and rules:** The natural logarithm function $$\ln(z)$$ is defined only for $$z > 0$$. Therefore, the argument inside the logarithm must be positive: $$\frac{\sin x - 1}{\sin x + 1} > 0$$ 3. **Analyze the inequality:** For a fraction to be positive, numerator and denominator must be both positive or both negative. - Numerator: $$\sin x - 1$$ - Denominator: $$\sin x + 1$$ 4. **Check numerator and denominator signs:** - $$\sin x - 1 > 0 \implies \sin x > 1$$, which is impossible since $$\sin x \leq 1$$. - $$\sin x - 1 < 0 \implies \sin x < 1$$ (always true except at $$\sin x = 1$$). - $$\sin x + 1 > 0 \implies \sin x > -1$$ (always true since $$\sin x \geq -1$$). - $$\sin x + 1 < 0 \implies \sin x < -1$$ (impossible). 5. **Conclusion on signs:** The numerator is always $$\leq 0$$ and denominator is always $$> 0$$ except at points where denominator is zero. 6. **Domain restrictions:** The denominator $$\sin x + 1 = 0$$ when $$\sin x = -1$$, which happens at $$x = \frac{3\pi}{2} + 2k\pi$$ for integers $$k$$. 7. **Evaluate the fraction:** Since numerator $$\sin x - 1 \leq 0$$ and denominator $$\sin x + 1 > 0$$, the fraction $$\frac{\sin x - 1}{\sin x + 1} \leq 0$$. 8. **Since the argument of $$\ln$$ must be positive, and the fraction is $$\leq 0$$, the function is undefined for all real $$x$$. **Final answer:** The function $$y = \ln\left(\frac{\sin x - 1}{\sin x + 1}\right)$$ has no real domain because the argument of the logarithm is never positive.