Question

Please help establish the identity.
1-[(sin^2v)/(1+cosv)]= cosv

Answer 1

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

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

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

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

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

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

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

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

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