证 因为A为锐角,所以cosA>0。tanA=b*tanB<==>(tanA)^2=b^2*(sinB)^2/(cosB)^2<==>(tanA)^2=b^2*(sinB)^2/[1-(sinB)^2]将sinB=sinA/a代入整理为:(b^2-1)*(sinA)^2=b^2-a^2<==>(b^2-1)*[1-(cosA)^2]=b^2-a^2< 展开
证 因为A为锐角,所以cosA>0。tanA=b*tanB<==>(tanA)^2=b^2*(sinB)^2/(cosB)^2<==>(tanA)^2=b^2*(sinB)^2/[1-(sinB)^2]将sinB=sinA/a代入整理为:(b^2-1)*(sinA)^2=b^2-a^2<==>(b^2-1)*[1-(cosA)^2]=b^2-a^2<==>(cosA)^2=(a^2-1)/(b^2-1)<==>cosA=√[(a^2-1)/(b^2-1)].tanA-1/sin2A=tanB <==>cos(2A-B)=0因为A,B均为锐角,所以2A-B=90°,即sin2A=cosB,故选A。 收起
