Here's the question you clicked on:
burhan101
How do i simplify this
\[\huge \frac{ 1+\frac{ sinx }{ cosx } }{ 1- \frac{ sinx }{ cosx } } =\]
\[\frac{ \sin x }{ \cos x } =\tan x\]
yea i am not allowed to use tan
i have to keep it only sin & cos that's why i simplified it originally from tan
\[ \frac{ 1+\frac{ \sin x }{ \cos x } }{ 1- \frac{ \sin x }{ \cos x } } =\frac{ \cos x+ \sin x }{\cos x- \sin x }\]
how did you do that ? :S
\[\frac{ 1+\frac{ \sin x }{ \cos x } }{ 1- \frac{ \sin x }{ \cos x } }\times1\] \[\frac{ 1+\frac{ \sin x }{ \cos x } }{ 1- \frac{ \sin x }{ \cos x } }\times\frac {\cos x}{\cos x}\] \[ =\frac{ \cos x+ \sin x }{\cos x- \sin x }\]\]