Question

三角恒等变换题1

已知 tanx+1/tanx=5
求 (sin2x+cos2x-1)/(1-tanx)
不是光要答案,
要过程或解析
详细点

Answer 1

tanx+1/tanx
=sinx/cosx+cosx/sinx
=[(sinx)^2+(cosx)^2]/(sinxcosx) (分母用倍角公式)
=1/(1/2sin2x)
=2/sin2x
=5
所以 sin2x=2/5.
(sin2x+cos2x-1)/(1-tanx) (分母用倍角公式：cos2x-1=-2(sinx)^2)
=(2sinxcosx-2(sinx)^2)/(1-sinx/cosx)
=2sinxcosx(cosx-sinx)/(cosx-sinx)
=2sinxcosx
=sin2x
=2/5
即原式 = 2/5.