1-tan^2x 1 tan^2x 周期

tan^2x+1=sin^2x/cos^2x+1=(sin^2x+cos^2x)/cos^2x=1/cos^2x=sec^2x

1.左＝tan(x/2+π/4)+tan(x/2-π/4)=tan[(x/2+π/4)+(x/2-π/4)][1-tan(x/2-π/4)tan(x/2+π/4)]=tanx[1-(-1)]=2tan

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

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

令a=tanx,a属于Ry=(a^2-a+1)/(a^2+a+1)ya^2+ya+y=a^2-a+1(y-1)a^2+(y+1)a+(y-1)=0判别式大于等于0(y+1)^2-4(y-1)^2>=0

令a=tanx则a属于Ry=f(x)=(a-a+1)/(a+a+1)ya+ya+y=a-a+1(y-1)a+(y+1)a+(y-1)=0a是实数则方程有解所以判别式大于等于0(y+1)-4(y-1)>

不好意思,现在才看到.第四题不完整.前三题如图：

右边=2tanx/(1+sin²x/cos²x)=2tanx/[(cos²x+sin²x)/cos²x]=2tanx*cos²x=2sinx

∫(tan²x+tan⁴x)dx=∫tan²x(1+tan²x)dx=∫tan²xsec²xdx=∫tan²xdtanx=(1/

楼主题目错了.应该是2tan(x)=tan^2(x/2)+1吧.用用tanx=sinx/cosxsinx=2sin(x/2)cos(x/2)cosx=2cos^2(x/2)-1基本就可以解决了吧.

secx-tanx化简成2/(1+tan(x/2))?这个显然不成立.利用特殊值即可判断取x=0,则secx-tanx=1-0=0而2/(1+tan(x/2))=2/(1+0)=2∴secx-tanx

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

设y=(tan^2x-tanx+1)/(tan^2x+tanx+1)另tanx=a,则a属于Ry=(a^2-a+1)/(a^2+a+1)[a属于R]而a^2+a+1>0恒成立则由判别式法有;y(a^2

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

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

因为f(-1007)=tan(-1006)+tan(-1005)+.+tan(-1)+tan(0)+tan(1)+...tan(1005)+tan(1006)=0f(-1007x2-x)=f(-201

cosx/√[1+tan^2(x)]+sinx/√[1=cot^2(x)=-1.cosx/|secx|+sinx/|cscx|=-1.cosx/cosx-sinx/sinx=-1.【暂时假定x在第四象

就化简一下就可以了lim(x→0)[(2x)/(1+x^2)]/(secxtanx+sinx)=lim(x→0)[(2x)/(1+x^2)]/(sinx/cosx^2+sinx)=lim(x→0)2/

1.y=(tan^2x-csc^2x)/(tan^2x+cot^2x-1)=(sin^2x/cos^2x-1/sin^2x)/(sin^2x/cos^2x+cos^2x/sin^2x-1)=(sin^