Question

prove that : cos3a-cos2a+cosa/sin3a - sin2a+sina= cot2a

Answer 1

=Cos3A-Cos2A+CosA/Sin3A-Sin2A+SinA

=Cos3A+CosA-Cos2A/Sin3A+SinA-Sin2A

Let 3A=C and A=D

CosC+CosD=2×(cosA+B/2)×(cosA-B/2)

SinC+SinD=2×(sinA+B/2)×(CosA-B/2)

=(2×cos3A+A/2×cos3A-A/2)-Cos2A/2×(Sin3A+A/2)×(Cos3A-A/2)-Sin2A

=(2×cos4A/2×cos2A/2)-Cos2A/(2×Sin4A/2×cos2A/2)-Sin2A

=2×Cos2A×CosA-Cos2A/2×Sin2A×CosA-2Sin2A

Taking Cos2A common on numerator and Taking Sin2A common on denominator

=Cos2A(2CosA-1)/Sin2A(2CosA-1)

Dividing Cos2A-1

=Cos2A/Sin2A

=Cot2A[CosA/SinA=CotA]