已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1

问题描述:

已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1

sin^2B=2sin^2A-1求证tan^2A=2tan^B+1sin^2B=2sin^2A-1求证tan^2A=2tan^B+1

证明:sin^2B=2sin^2A-1得cos^2B=1-sin^2B=1-(2sin^2A-1)=2(1-sin^2A)=2cos^2A于是tan^2A-2tan^B=sin^2A/cos^2A-2sin^2B/cos^2B=sin^2A/cos^2A-2sin^2B/(2cos^2A)=sin^2A/cos^2A-sin^2B/cos^2A=(sin^2A-sin^2B)/cos^...