How to solve this identity? Thank you!

#(tan(a+b)+tan(a-b))/(tan(a+b)-tan(a-b)) = (sin(2a))/(sin(2b))#

1 Answer
Abhishek K.
Dec 10, 2017

We know that, #tana+-tanb#

#=sina/cosa+-sinb/cosb#

#=(sinacosb+-cosasinb)/(cosacosb#

#=sin(a+-b)/(cosa*cosb)#

So, #(tana+tanb)/(tana-tanb)=sin(a+b)/sin(a-b)#

#LHS=(tan(a+b)+tan(a-b))/(tan(a+b)-tan(a-b)#

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

#=sin(2a)/(sin(2b))=RHS#

Related questions
Impact of this question
1072 views around the world
You can reuse this answer
Creative Commons License