SOLUTION: What is sin(u+v)+sin(u-v)/cos(u+v)+cos(u-v)=tan (u) I have tried answering it: LHS>>> =[sin(u)cos(v)+cos(u)sin(v)]+[sin(u)cos(v)+cos(u)sin(v)]/[cos(u)cos(v)-sin(u)sin(v)]+[c

Algebra ->  Trigonometry-basics -> SOLUTION: What is sin(u+v)+sin(u-v)/cos(u+v)+cos(u-v)=tan (u) I have tried answering it: LHS>>> =[sin(u)cos(v)+cos(u)sin(v)]+[sin(u)cos(v)+cos(u)sin(v)]/[cos(u)cos(v)-sin(u)sin(v)]+[c      Log On


   

Question 889169: What is
sin(u+v)+sin(u-v)/cos(u+v)+cos(u-v)=tan (u)
I have tried answering it:
LHS>>>
=[sin(u)cos(v)+cos(u)sin(v)]+[sin(u)cos(v)+cos(u)sin(v)]/[cos(u)cos(v)-sin(u)sin(v)]+[cos(u)cos(v)+sin(u)sin(v)]
=sin(u)cos(v)+sin(u)cos(v)/cos(u)cos(v)+cos(u)cos(v)
=2sin(u)cos(v)/2cos(u)cos(v)
and i'm stuck here. I don't know how i'll make it equal to the RHS (tan u).
Please help me out. Thanks
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
It's right there.
%282sin%28u%29cos%28v%29%29%2F%282cos%28u%29cos%28v%29%29
Cancel out the 2cos%28v%29 terms from numerator and denominator.
sin%28u%29%2Fcos%28u%29
tan%28u%29