1. **State the problem:** Prove that $$\frac{\tan y}{\sin y} = \sec y$$. 2. **Recall definitions:** - $$\tan y = \frac{\sin y}{\cos y}$$ - $$\sec y = \frac{1}{\cos y}$$ 3. **Rewrite the left-hand side (LHS):** $$\frac{\tan y}{\sin y} = \frac{\frac{\sin y}{\cos y}}{\sin y}$$ 4. **Simplify the complex fraction:** $$= \frac{\sin y}{\cos y} \times \frac{1}{\sin y}$$ 5. **Cancel common factors:** $$= \frac{\cancel{\sin y}}{\cos y} \times \frac{1}{\cancel{\sin y}} = \frac{1}{\cos y}$$ 6. **Recognize the right-hand side (RHS):** $$\frac{1}{\cos y} = \sec y$$ 7. **Conclusion:** $$\frac{\tan y}{\sin y} = \sec y$$ is proven. This shows the identity holds true by expressing all terms in sine and cosine and simplifying step-by-step.