Prove that 4sinA.sin(60-A).sin(60+A)

Steps to prove 4sinA.sin(60-A).sin(60+A)-sin3A=0

Take L.H.S 4sinA.sin(60-A).sin(60+A)-sin3A

4sinA.sin(60-A).sin(60+A)-sin3A

[sinA.sin(60-A).sin(60+A)=1/4sin3A]

=4.1/4sin3A-sin3A

=sin3A-sin3A

=0=R.H.S

Hence 4sinA.sin(60-A).sin(60+A)-sin3A=0 is proved.

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