求证：1.sin^4x+cos^4x=1-2sin^2cos^2x2.1-tan^2x/1+tan^2x=cos^2x-

（1）sin^4 x+cos^4 x＝（sin^2 x+cos^2 x）^2－2sin^2 xcos^2 x =1-2sin^2 xcos^2 x
(2)(1-tan^2 x)/(1+tan^2x)=(cos^2 x/cos^2 x-sin^2 x/cos^2 x)/)=(cos^2 x/cos^2 x+sin^2 x/cos^2 x)=[(cos^2 x-sin^2 x )/cos^2 x×cos^2 x/(cos^2 x)sin^2 x )=cos^2 x-sin^2 x
(3)因为tanx=-1/3,所以：cosx≠0
把（7sinx-3cosx）/（4sinx+5cosx）分子、分母同时除以cosx,得：
（7sinx-3cosx）/（4sinx+5cosx）＝（7sinx/cosx-3cosx/cosx）/（4sinx/cosx+5cosx/cosx）=(7tanx-3)/(4tanx+5)=[7×(-1/3)-3]/[4×(-1/3)+5]=-16/11
注意：tanx=sinx/cosx

3.分子分母同时除以sinx得7-3tanx/4+5tanx
tanx=-1/3
原式=40/17

SIN^4X+COS^4X=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1-(2sinxcosx)^2/2

sin^4x+cos^4x
=(sin^2x)^2 +(cos^2x)^2 +2sin^2xcos^2x -2sin^2xcos^2x
=(sin^2x+cos^2x)-2sin^2xcos^2x
=1-2sin^2xcos^2x
1-tan^2x/1+tan^2x
=(1-sin^2x/cos^2x)/(1+sin^2x/cos^2x)
=(cos...