若tan^2x-sin^2x 16 5

左边=(1-sin²x/cos²x)/(1+sin²x/cos²x)上下乘cos²x=(cos²x-sin²x)/(cos

用图象来证明,不好说啊总之这个是凹函数的一个性质,你把tanX的图象的(0,pi/2)部分的图象画出来,然后随便在(0,pi/2)内取两个点令X1,X2,把X1,X2对应的函数值在图象上标出来然后给两

是[1-(tanx)^2]/[1+(tanx)^2]=(cosx)^2-(sinx)^2=========证明：[1-(tanx)^2]/[1+(tanx)^2]={[1-(tanx)^2]*(cos

tan²-1=sin²x/cos²x-1=(sin²x-cos²x)/cos²x=(sinx+cosx)(sinx-cosx)/cos&su

不好意思,现在才看到.第四题不完整.前三题如图：

sin(π-x)+sin(π+x)-cos(-x)+cos(2π-x)-tan(π+x)cot(π-x)=-sin(-x)-sin（x）-cos(x)+cos(-x)-[-tan(x)][-cot(-

由已知2sin(4x1-π/3)-a=2sin(4x2-π/3)-a=0也就是sin(4x1-π/3)=sin(4x2-π/3)=a/2所以说（4x1-π/3）+（4x2-π/3）=kπ(k是整数)或

sin(2派-x)tan(派+x)cot(-x-派)/tan(3派-x)cos(派-x)=-sinxtanxcot(-x)/tan(-x)(-cosx)=sinxtanxcotx/tanxcosx=s

tan,正切；sin,正弦；cos,余弦tan(x+y)tan(x-y)=sin(x+y)/cos(x+y)*sin(x-y)/cos(x-y)=sin(x+y)sin(x-y)/[cos(x+y)c

sin(tt/2+x)*cos(tt/2-x)*tan(-x+3tt)/sin(7tt-x)*tan(6tt-x)=-cosx*sinx*(-tanx)/sinx*(-tanx)=-cosx*tanx

sin^2x/(sinx-cosx)-(sinx+cosx)/(tan^2x-1)=sin^2x/(sinx-cosx)-(sinx+cosx)/[(tanx+1)(tanx-1)]=sin^2x/(

(1)f(-x)=sinacosx-(tana-2)sinx-sina=f(x),则tana-2=0,那么tana=2(2)f(x)=sinacosx-sina=sina(cosx-1)f(x)>=0

tan^2x-sin^2x=sin^2x/cos^2x-sin^2x=(1/cos^2x-1)sin^2x=[(1-cos^2x)/cos^2x]sin^2x=[sin^2x/cos^2x]sin^2

(1/sin^2x)+(1/cos^2x)-(1/tan^2x)=[(cosx)^2+(sinx)^2]/(sinxcosx)^2-(cosx)^2/(sinx)^2=1/(sinxcosx)^2-(

解由sinx-cosx=√2（√2/2sinx-√2/2cosx）=√2（cos45°sinx-sin45°cosx）=√2sin（x-45°）=根号2sin(x+α),即α=-45°即tanα=ta

左边=(1-2sinxcosx)/(cos²x-sin²x)=(sin²x+cos²x-2sinxcosx)/(cos²x-sin²x)=(

x=(-4*pi:0.1:4*pi);y=tan(sin(x.^2));plot(x,y)这是从-4pi到4pi的图像.

因为sin最值是-1和1所以f(x1)

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)

1(1)⑴[Cos(x—π)tan(x—2π)tan(2π—x)]/[sin(π+x)]=cos(π-x)tanx(-tanx)/(-sinx)=-cosxtanx(-tanx)/(-sinx)=-s