Question

Prove:
Cos3A.Sin9A - SinA.Cos5A/CosA.cos5A - Sin3A.Sin9A = tan8A

Answer 1

Answer:

tan8A

Step-by-step explanation:

L. H. S=cos3A. sin9A-sinA. cos5A/cosA. cos5A -sin3A. sin9A

formula used_2cosAsinB=sin(A+B)-sin(A-B)

=1/2[2cos3A.sin9A-sinA.cos5A/1/2[2cosA.cos5A - sin3A. sin9A

cancellation of 1/2 and 2

=sin(3A+9A)-sin(3A-9A)-sin(A+5A)+sin(A+5A)/cos(A+5A)

+cos(A-5A)-cos(3A-9A)+cos(3A+9A)

=sin12A-sin(-6A)-sin6A-sin(-4A)/cos6A+cos(-4A)-cos(-6A)+cos12A

cancellation of sin6A in numerator and cos4A in denominator

=sin12A-sin4A/cos12A-cos4A

=sin8A/cos8A

=tan8A

=R. H. S