[tex]\begin{gathered} We\text{ are asked to simplify the below equation:}\frac{}{} \\ \frac{sin^2y}{1-cosy} \\ Using\text{ trigonometric identities we can re-write it as:} \\ \frac{1-cos^2y}{1-cosy} \\ The\text{ numerator is a difference of squares} \\ =\text{ }\frac{(1+cosy)(1-cosy)}{(1-cosy)} \\ =\text{ 1+ cosy} \end{gathered}[/tex]
Answer: 1 + cosy