求证tan^2x-sin^2x=tan^2x*sin^2x

左边=(sin^2x+cos^2x+2sinxcosx)/(cosx+sinx)(cosx-sinx)=(cosx+sinx)^2/(cosx+sinx)(cosx-sinx)=(cosx+sinx)

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

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

sin(x+y)=1所以x+y=2kπ+π/2所以2x+y=2(x+y)-y=4kπ+π-y所以tan(2x+y)=tan(4kπ+π-y)因为tan的周期是π所以tan(4kπ+π-y)=tan[4

左边=sin²x/(sinx-cosx)-(sinx+cosx)/(sin²x/cos²x-1)=sin²x(sinx+cosx)/(sinx-cosx)(si

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

你看后面TAN里一个x一个x+y那你就把给你的原式中的2x+y拆开,在消消化化的,试下吧我觉得能行

tan（2x＋β）=(tanx+tan(2x＋β))/(1-tanxtan(2x＋β))=(tanx+1)/(1-tanx)=-cos/sinx故tan（2x＋β）＋cosx/sinx＝0

令a=x+y,则条件变为3sin(a-x)=sin(a+x),展开得3sinacosx-3cosasinx=sinacosx+cosasinx,移项2sinacosx=4cosasinxtana=2t

已知sin（x+y）=1,求证：tan（2x+3y）=tany证明：sin（x+y）=1所以x+y=2k兀+兀/2K为整数所以tan(2x+3y)=tan(4k兀+兀+y)=tan(兀+y)=tany

sin(2α+β)=sin(2α)cosβ+cos(2α)sinβ=3sinβsin(2α)+cos(2α)tanβ=3tanβ[3-cos(2α)]tanβ=sin(2α)tanβ=sin(2α)/

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

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

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

sin[(x+y)+x]=5sin[(x+y)-x]sin(x+y)·cosx＋cos(x+y)·sinx＝5·sin(x+y)·cosx－5·cos(x+y)·sinx4·sin(x+y)·cosx

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

第一题：（1＋2sinxcosx）/［（cosx）^2－（sinx）^2］＝［（cosx）^2＋2sinxcosx＋（sinx）^2］/［（cosx）^2－（sinx）^2］＝（cosx＋sinx）^

由sin(x+y)=1可知x+y=90度tan(2x+y)+tanytan(2x+y+y)tan(2x+2y)tan180度因为:tan180度=0(常识!)所以:tan(2x+y)+tany=0

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