Question

sina-sin2a-sin4a+sin5a/cosa-cos2a-cos4a+cos5a

Answer 1

cosA+cosB=2cos(A+B/2)cos(A-B/2) and sinA+sinB=
2sin(A+B/2)cos(A-B/2). sina+sin5a-(sin2a+sin4a)/
cosa+cos5a-(cos2a+cos4a) =
2sin(a+5a/2)cos(a-5a/2)-2sin(2a+4a/2)cos(2a-4a/2)
/ 2cos(5a+a/2)cos(a-5a/2)-2cos(2a+4a/2)cos(2a-4a/2) =2sin(3a)cos(2a)-2sin(3a)cos(a)/2cos(3a)cos(2a) -
2cos(3a)cos(a)= sin(3a)(cos(2a)-cos(a))/cos(3a)(cos(2a)-cos(a))= sin(3a)/cos(3a) =tan(3a).
[cos(x)=cos(-x)] .