identities: sin^2/cosx = secx- cosx

[ca.chick]

sin^2/cosx = secx- cosx

i dont know what side to simplfy, and how there is a squared on one side and not the other

 

[morson]

LHS is:

\(\displaystyle \frac{sin^2x}{cosx}\ = sin^2x \frac{1}{cosx}\ = sin^2x secx = (1 - cos^2x) secx , cosx \not=\ 0\)

Finish it off.

 

[soroban]

Re: identities

Hello, ca.chick!

Do you know any of the basic identities?

\(\displaystyle \L\frac{\sin^2x}{\cos x} \:= \:\sec x\,-\,\cos x\)

The right side is: \(\displaystyle \L\:\sec x\,-\,\cos x \;=\;\frac{1}{\cos x}\,-\,\cos x \;=\;\frac{1\,-\,\cos^2x}{\cos} \;=\;\frac{\sin^2x}{\cos x}\)