[(1/2)sinB/cosB]/[1+(3/2)sin²Bcos²B]
=sinB/[2cosB+3(sinB)^2*(cosB)^3],(1)
sinBcosB/2cos²B+3sin²B
=sinB/(2cosB)+3(sinB)^2
=sinB(1+6sinBcosB)/(2cosB).(2)
(1)=(2)
1+6sinBcosB=1+(3/2)(sinB)^2*(cosB)^2,
sinBcosB=4,不可能.
[(1/2)sinB/cosB]/[1+(3/2)sin²Bcos²B]
=sinB/[2cosB+3(sinB)^2*(cosB)^3],(1)
sinBcosB/2cos²B+3sin²B
=sinB/(2cosB)+3(sinB)^2
=sinB(1+6sinBcosB)/(2cosB).(2)
(1)=(2)
1+6sinBcosB=1+(3/2)(sinB)^2*(cosB)^2,
sinBcosB=4,不可能.