1. The problem involves two expressions:\n$$\frac{1}{1} + \frac{1}{\text{something}} = 1$$\nand\n$$\frac{1 + \sin^2 A}{1 + \csc^2 A}$$\nWe need to analyze and simplify these.\n\n2. First expression: $$\frac{1}{1} + \frac{1}{1} = 1$$ is incorrect, because $$1 + 1 \neq 1$$. Possibly the question intended fractional expressions added to equal 1, but not clear.\n\n3. Second expression: Simplify $$\frac{1 + \sin^2 A}{1 + \csc^2 A}$$\nRecall $$\csc A = \frac{1}{\sin A}$$ so $$\csc^2 A = \frac{1}{\sin^2 A}$$\n\n4. Substitute: $$\frac{1 + \sin^2 A}{1 + \frac{1}{\sin^2 A}} = \frac{1 + \sin^2 A}{\frac{\sin^2 A + 1}{\sin^2 A}}$$\n\n5. Simplify the denominator by multiplying numerator and denominator: $$= \left(1 + \sin^2 A\right) \times \frac{\sin^2 A}{\sin^2 A + 1}$$\n\n6. Note that $$1 + \sin^2 A = \sin^2 A + 1$$ so numerator and denominator are the same expression, hence cancel:\n$$= \sin^2 A$$\n\n7. Therefore, $$\frac{1 + \sin^2 A}{1 + \csc^2 A} = \sin^2 A$$\n\n8. Interpretation based on the right triangle where hypotenuse = 1, opposite side = $$\sin A$$, adjacent side = $$\cos A$$:\n\- $$\sin A = \text{opposite} / \text{hypotenuse} = \sin A / 1 = \sin A$$\n\- So the result $$\sin^2 A$$ matches the square of the opposite side over the hypotenuse squared, confirming the expression geometrically.\n\nFinal Answer: $$\boxed{\sin^2 A}$$