tan(a+b) identity derivation?

Question

Hi
I am trying to derive tan(a+b)
I have done:
tan(a+b)=sin(a+b)/cos(a+b)=sinAcosB+cosAsinB / cosAcosB-sinAsinB
can someone tell me where to go next?

Thanks!

the part of the sin (a+b) expansion that is dotted out says:

sin A cos B + cos A sin B

Answer 1

OK, so you have
sin(a+b)/cos(a+b)

= (sinAcosB+cosAsinB)/(cosAcosB-sinAsinB)

Divide each term on both sides of the fraction by cosAcosB:

you get


(tanA*1+1*tanB)/(1-tanAtanB)

= (tanA+tanB)/(1-tanAtanB)

Answer 2

You know that
tan(a+b)=(sinAcosB+cosAsinB)/(cosAcosB-sinAsinB)
=divide denominator and numerator with cosAcosB
you will get
=(sinA/cosA+sinB/cosB)/(1-sinAsinB/(cosAcosB))
=that in term turns into tan function
=(tanA+tanB)/(1-tanAtanB)

Answer 3

tan(A + B)
= sin(A + B)/cos(A + B)
= (sinAcosB + cosAsinB)/(cosAcosB-sinAsinB)
Now divide numerator and denominator by cosAcosB
So tan(A + B)
= (sinA/cosA + sinB/cosB)/(1 - (sinA/cosA)*(sinB/cosB))
= (tanA + tanB)/(1 - tanAtanB)

Answer 4

It's quite easy you known: (sin(A-B))/(sin(A+B)) = (sinAcosB- cosAsinB)/ (sinAcosB + cosAsinB) +) you only divide both numerator and denominator with cosAcosB you have ((sinAcosB - cosAsinB)/ (cosAcosB)) /((sinAcosB + cosAsinB)/ (cosAcosB)) = (tanA-tanB)/(tanA+tanB)