Question

1-(cos^2x)/(1+sinx) write as a single trig function

Answer 1

\[\frac{1-\cos^2 x}{1+\sin x} = \frac{\sin^2 x}{1+\sin x}\]
multiply by conjugate
\[\frac{\sin^2 x}{1+\sin x}*\frac{1-\sin x}{1-\sin x} = \frac{\sin^2 x (1-\sin x)}{1-\sin^2 x} = \frac{\sin^2 x(1-\sin x)}{\cos^2 x}\]
\[= \tan^2 x (1-\sin x)\]

hmm thats as far as i can go i think