高一数学 三角函数

Question

如题所述

Answer 1

To find log<tanx> cosx = ?
log<tanx+cotx> sinx = -3/4
log<tanx> sinx/log<tanx>( tanx + cotx ) = -3/4
log<tanx>( tanx.cosx) /log<tanx>{ [(tanx)^2+1]/tanx } = -3/4
(1+ log<tanx> cosx) / log<tanx>{ 1/[tanx . (cosx)^2 ] } = -3/4
(1+ log<tanx> cosx) / ( -1-2log<tanx>cosx ) = -3/4
4(1+ log<tanx> cosx) = 3( 1+2log<tanx>cosx )
2log<tanx> cosx = 1
log<tanx> cosx = 1/2

Answer 2

如图

追答

Answer 3

追答

看错题了，上边的作废。