How do you rewrite #(tan20+tan32)/(1-tan20tan32)# as a function of a single angle and then evaluate?

Question

2547 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-27T10:34:17

The answer is #=tan52=1.28#

We need

#sin(a+b)=sinacosb+sinbcosa#

#cos(a+b)=cosacosb+sinasinb#

#(tana+tanb)/(1-tanatanb)=(sina/cosa+sinb/cosb)/(1-sina/cosa*sinb/cosb)#

#=((sinacosb+sinbcosa)/(cosacosb))/((cosacosb-sinasinb)/(cosacosb))#

#=(sinacosb+sinbcosa)/(cosacosb-sinasinb)#

#=sin(a+b)/cos(a+b)#

#=tan(a+b)#

Therefore,

#(tan20+tan32)/(1-tan20tan32)=tan(20+32)=tan52=1.28#