Here's the question you clicked on:
Shaydenparis
prove siny +tany/over/ 1+secy=siny
\(\dfrac{\sin y + \tan y}{1 + \sec y} \Rightarrow \dfrac{\sin y + \tan y}{1 + \sec y} \cdot \dfrac{1 - \sec y}{1 - \sec y} = \dfrac{(\sin y + \tan y)(1 - \sec y)}{1 - \sec^2 y} \) Can you do the next step?
yep i got it, thank you
For some reason, I don't really see how this helps... better to express everything in terms of sin and cos instead... oh well.
\[\huge \frac{\sin y + \frac{\sin y}{\cos y}}{1 + \frac{1}{\cos y}}\]
\[\huge \frac{\frac{\sin y \cos y + \sin y}{\cos y}}{\frac{\cos y + 1}{\cos y}}\]
\[\large \frac{\sin y \cos y + \sin y}{\cos y + 1}=\frac{(\sin y)(\cos y + 1)}{\cos y + 1}\] And now it should be easy to see.