已知sin^2B=2sin^2A-1求证tan^2A=2tan^B+1

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

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

已知sin^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^2A=[sin^2A-(2sin^2A-1)]/cos^2A
=(1-sin^2A)/cos^2A=1
故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