Question

(sin^2x-2sinx*cosx-cos^2x)/(4cos^2x-3sin2x），已知tanx=2，求值

Answer 1

(sin^2x-2sinx*cosx-cos^2x)/(4cos^2x-3sin2x)
=(sin^2x-2sinx*cosx-cos^2x)/(4cos^2x-6sinxcosx)分子分母同时除以cos^2x
=(sin^2x/cos^2x-2sinx*cosx/cos^2x-cos^2x/cos^2x)/(4cos^2x/cos^2x-6sinxcosx/cos^2x)
=(tan^2x-2tanx-1)/(4-6tanx)
=(2^2-2*2-1)/(4-6*2)
=-1/(-8)
=1/8

Answer 2

原式=[sin²x－2sinacosx－cos²x]/[4cos²x－6sinxcosx]
=[tan²x－2tanx－1]/[4－6tanx] 【分子分母同除以cos²x】
=[2²－2×2－1]/[4－6×2]
=1/8