2sinAcos3A-2sin3AcosA=
sin4A
12sin4A
14sin4A
none of these
Explanation for correct option:
Given, 2sinAcos3A-2sin3AcosA
=2cosAsinAcos2A-sin2A
Apply formula, sin(2x)=2sin(x)cos(x)&cos(2x)=cos2(x)-sin2(x)
β2cosAsinAcos2A-sin2A=sin2Acos2A=122sin2Acos2A=12sin4A
Therefore, correct answer is option B