Question

So, tan(a+b)=sin(a+b)/cos(a+b)

I need to simplify and the answer is:

tan(a)+tan(b)/1-tan(a)tan(b)

I understand all the identities and division but where does the 1 in the denominator come from?

Answer 1

expand sin(a+b) and cos (a+b)
what u get ?

Answer 2

Do you know the respective formulae for sin(a + b) and cos(a + b)?

Answer 3

tanx=sinx/cosx
cotx=cosx/sinx

Answer 4

I end up with sin(a)cos(b)+cos(a)sin(b)/cos(a)cos(b)-sin(a)sin(b)

Answer 5

Then if I divide each term by cos(a)cos(b), that's where I get tan(a)+tan(b)/tan(a)tan(b) but the teacher says the denomenator has a 1 in it???

Answer 6

\[\tan(a+b)=\frac{ \sin(a+b) }{ \cos(a+b) }\]
\[=\frac{ \sin(a)\cos(b) +\cos(a)\sin(b)}{ \cos(a)\cos(b)-\sin(a)\sin(b) }\]\[=\frac{ \ \frac{ \sin(a)\cos(b) }{ \cos(a)\cos(b) }+\ \frac{ \cos(a)\sin(b }{ \cos(a)\cos(b) })}{ \frac{ \cos(a)\cos(b) }{ \cos(a)\cos(b) }-\frac{ \sin(a)\sin(b) }{ \cos(a)\cos(b) }}\]\[=\frac{ \tan(a)+\tan(b) }{ 1-\tan(a)\tan(b) }\]

Answer 7

@Zekarias

Answer 8

thanks

Answer 9

Ugh...I see it. Thanks guys!