prove that tan(a+b)tan(a-b) = sin2a-sin2b/cos2a-sin2b Share with your friends Share 1 Ajanta Trivedi answered this LHS of the given equation is: tan(a+b).tan(a-b)=sin(a+b)cos(a+b).sin(a-b)cos(a-b)=2.sin(a+b).sin(a-b)2cos(a+b).cos(a-b) [multiplying the Nr and Dr by 2=cos2b-cos2acos2a+cos2b [since 2cosAcosB=cos(A+B)+cos(A-B) and 2sinA.sinB=cos(A-B)-cos(A+B)=1-2sin2b-(1-2sin2a)2cos2a-1+1-2sin2b [since cos2θ=2cos2θ-1=1-2sin2θ=2sin2a-2sin2b2cos2a-2sin2b=sin2a-sin2bcos2a-sin2b =RHS hope this helps you 1 View Full Answer
