Question

prove 1-(cos^2x/sinx+1) = sin x

Answer 1

\[
1-\frac{\cos^2x}{\sin x+1}=\sin x\\
(\sin x+1)-\cos^2x=\sin x(\sin x+1)\\
\sin x+1-\cos^2x=\sin^2x+\sin x\\
1-\cos^2x=\sin^2x\\
\boxed{\sin^2x+\cos^2x=1}
\]

Answer 2

In case it isn't clear, the step to get from the first line to the second line is to multiply both sides by \((\sin x+1)\).

Answer 3

you cannot change sinx on the right.. that is why it is a proof

Answer 4

to prove a proof, u need both sides to say the exact same, in order to do that, u can change the other side to simplify the expression on the left

Answer 5

\[1-\frac{\cos ^2x}{\sin x+1}=\frac{\sin x+1-(1-\sin ^2x)}{\sin x+1}=\frac{\sin ^2x+\sin x}{\sin x+1}=\]

Answer 6

\[LS = 1-(\frac{cos^2x}{sinx+1})\]\[ = \frac{sinx+1}{sinx+1}-\frac{1-sin^2x}{sinx+1}\]\[= \frac{sinx+1-1+sin^2x}{sinx+1}\]\[= \frac{sinx+sin^2x}{sinx+1}\]\[ = \frac{sinx(sinx+1)}{sinx+1}\]\[ = sinx\]\[ =RS\]

Answer 7

\[\frac{\sin x(\sin x+1)}{\sin x+1}=\sin x\]