Question

cot-tan/sin+cos= csc-sec

Answer 1

come here to see http://openstudy.com/study#/updates/526877e6e4b040e33d2ef83e

Answer 2

\[cotx-\frac{tanx}{sinx}+cosx=cscx-secx\]
@alejandrop95 do you still need help with this?

Answer 3

tanx= sinx/cosx
so\[\frac{cosx}{sinx}+\frac{sinx}{cosxsinx}+cosx=\frac{1}{sinx}-\frac{1}{cosx}\]
\[\frac{cosx}{sinx}+\frac{1}{cosx}+cosx=\frac{1}{sinx}-\frac{1}{cosx}\]
\[\frac{cosx}{sinx}+cosx=\frac{1}{sinx}-\frac{2}{cosx}\]
\[\frac{cosx}{sinx}-\frac{1}{sinx}+cosx=-\frac{2}{cosx}\]
\[\frac{cosx-1}{sinx}+cosx=-\frac{2}{cosx}\]

Answer 4

\[\frac{cosx-1}{sinx}+\frac{cosx}{1}=\frac{2}{cosx}\]
\[\frac{cosx-1}{sinx}+\frac{cosxsinx}{1sinx}=\frac{2}{cosx}\]
\[\frac{cosx-1}{sinx}+\frac{cosxsinx}{sinx}=\frac{2}{cosx}\]
\[\frac{cosx-1+cosxsinx}{sinx}=\frac{2}{cosx}\]
\[(cosx-1+cosxsinx)cosx=2sinx\]
I cross multiplied.
\[(\cos^2x-cosx+\cos^2xsinx)=2sinx\]

can you do it from here?