Proof ? [1 + tan2θtan3θ]/[1 + tanθtan3θ]=cos^2θ/cos^2θ

#(1 + tan2θtan3θ)/(1+ tanθtan3θ]=cos^2θ/cos^2 (2θ)#

1 Answer
Feb 28, 2018

#LHS=(1 + tan2θtan3θ)/(1+ tanθtan3θ)#

#=(1 + (sin2θsin3θ)/(cos2thetacos3theta))/(1+ (sinθsin3θ)/(costheta cos3theta))#

#=((cos3thetacos2theta + sin2θsin3θ)/(cos2thetacos3theta))/((cos3thetacostheta+ sinθsin3θ)/(costheta cos3theta))#

#=((cos(3theta-2theta))/(cos2thetacos3theta))/((cos(3theta-theta))/(costheta cos3theta))#

#=((costheta)/(cos2thetacos3theta))/((cos(2theta))/(costheta cos3theta))#

#=cos^2θ/cos^2 (2θ)=RHS#