## anonymous 3 years ago How do i simplify this

1. anonymous

$\huge \frac{ 1+\frac{ sinx }{ cosx } }{ 1- \frac{ sinx }{ cosx } } =$

2. UnkleRhaukus

$\frac{ \sin x }{ \cos x } =\tan x$

3. anonymous

yea i am not allowed to use tan

4. anonymous

i have to keep it only sin & cos that's why i simplified it originally from tan

5. UnkleRhaukus

$\frac{ 1+\frac{ \sin x }{ \cos x } }{ 1- \frac{ \sin x }{ \cos x } } =\frac{ \cos x+ \sin x }{\cos x- \sin x }$

6. anonymous

how did you do that ? :S

7. UnkleRhaukus

$\frac{ 1+\frac{ \sin x }{ \cos x } }{ 1- \frac{ \sin x }{ \cos x } }\times1$ $\frac{ 1+\frac{ \sin x }{ \cos x } }{ 1- \frac{ \sin x }{ \cos x } }\times\frac {\cos x}{\cos x}$ $=\frac{ \cos x+ \sin x }{\cos x- \sin x }$\]