y=1 tan^2x-2tanx 2的值域

设y－x＝ttan(y-x)=-1/7=>tant=-1/7tan(y-2x)=tan(t+x)=(tant+tanx)/(1-tant*tanx)=(2-1/7)/(1+2/7)=13/9

解;首先求定义域:tanx+根号(1+tan^2x)>01+tan^2x>tan^2x恒成立则x也可以取任意数,关于原点对称f(x)=lg(tanx+根号(1+tan^2x)f(-x)=lg(tan(

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)>

设t=tanx,t∈R则y=1/(t^2-2t+2)y=1/[(t-1)^2+1]∵(t-1)^2+1≥1∴1/[(t-1)^2+1]∈(0,1]即y=1\(tan^2x-2tanx+2)的值域为(0

tanx属于实数R设t=tanxy=t^2+t+1=(t+1/2)^2+3/4所以函数值域是[3/4,正无穷）tanx=1,(-TT/2,TT/2)和(0,2TT）X的值(-TT/2,TT/2)里,x

(1)求函数y=(tan²x-tanx+1)/(tan²x+tanx+1)的值域由原式得ytan²x+ytanx+y=tan²x-tanx+1故有(y-1)ta

sin(x+y)=sinxcosy+cosxsiny=1/2sin(x-y)=sinxconsy-cosxsiny=1/3sinxcosy=5/12,cosxsiny=1/12tanx/tany=si

tanx+cotx=1/sinxcosx=2/sin2x

点小图看大图

x不等于5/6π+kπx不等于π/4+kπ并且不等于π/2+kππ/2+kπ>x>-π/3+kπ

y=tanx/1-tan^2x=1/2tan2x,周期即π/2记得采纳啊

定义域为tanx的定义为x不等于（2k+1）∏/2值域y=4x/(x~2+1)当x=0时y=0当x不等于0时y=4/(x+1/x)此时得到y=-2故他的值域为-2〈=y

y=1/(tan²x-2tanx+2)=1/[(tanx-1)^2+1]因此最大值1,值域(0,1]定义域x≠kπ+π/2再问：能不能再详细点。。再答：我配方的不不清楚吗？再问：谢了。。懂了

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

令a=tanx则a∈Ry(a²+a+1)=a²-a+1(y-1)a²+(y+1)a+(y-1)=0a是实数则方程有解判别式大于等于0(y+1)²-4(y-1)&

令a=tanx则a属于Ry=f(x)=(a²-a+1)/(a²+a+1)ya²+ya+y=a²-a+1(y-1)a²+(y+1)a+(y-1)=0a是

tan(2x-y)=[tanx+tan(x-y)]/[1-tanxtan(x-y)]=(1/10)/(1+1/5）=1/12

y=sinx(1+tanx*tan(x/2))=sinx{1+(sinx/cosx)*[(1-cosx)/sinx]}=sinx[1+1/cosx-1]=sinx/cosx=tgx.所以最小周期是tg

tan(x+y)=(tanx+tany)/(1-tanxtany)0.4=（tany+0.5)/(1-0.5*tany)就可以解出来了.希望对你有所帮助!

令x/2=a,则x=2a所以tan2a=2tana/(1-tan²a)=2×2/(1-2²)=4/(-3)=-4/3即tanx=-4/3再问：为什么tanx的答案不换为2a呢？再答