tanπ2x

(tanx/2-π/6)

tan(π/2-x),tan(π/2+x),tan(x)有什么关系tan(π/2-x)=1/tan(x)tan(π/2+x)=-1/tan(x)tan(π/2-x)=-tan(π/2+x),

极限(1-x)tanπx/2

lim(2-x)tanπ/4x应用罗比达法则lim(2-x)tanπx/4=lim(2-x)/cot(πx/4)=lim(-1/[-πcsc^2(πx/4)/4])=4limsin^2(πx/4)/π=4/π

设f(x)=(tan(π/4)x-1)(tan(π/4)x²-2).(tan(π/4)x^100-100),求f'(1)tan(π/4)=1f(x)=(x-1)(x^2-2)...(x^100-100)y=uv,y'=u'v+uv

tan（x/2+π/4）+tan（x/2-π/4）=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1

证明下列恒等式(1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx(1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx（2）（1-2sinαcosα/cos²α-sin²α）=（1-tan

y=tan(π/2-x)定义域y=tan(π/2-x)=cotx故定义域为{x|x≠kπ,k∈Z}π/2-x≠kπ+π/2,x≠kπ

tanx+tan(π/2-x)=?tanx+cotx=1/sinxcosx=2/sin2x

满足不等式tan(2x+π/3)tan(2x+π/3)≤1所以kπ-π/2＜2x+π/3≤kπ+π/4,k∈Z所以kπ-5π/6＜2x≤kπ-π/12,k∈Z故kπ/2-5π/12＜x≤kπ/2-π/24,k∈Z即x的取值范围是(kπ/2-

tan(-x+π/4)tan(-x+π/4)tan(-x+π/4)-tan(x-π/4)tan(x-π/4)>0由图像知，kπ解集为（kπ+π/4，kπ+3π/4）k∈z

(tanπx/4)^(tanπx/2)当x趋向于1时的极限?此为1^(∞)型,经过适当变形化简,再用lim(1+x)^(1/x)=ex→0原极限=lim[1+(tanπx/4-1)]^{[1/(tanπx/4-1)]▪tanπx

x→1,求lim（tanπx/4）^tanπx/2求极限,lim（tanπx/4）^tanπx/2=lim1^tanπx/2=1

求极限.(tanx)^(tan2x)x→π/4这种先令y=(tanx)^(tan2x)然后求对数lny=tan2x(ln(tanx))=ln(tanx)/cot2xx->π/40/0洛必达=(1/tanx)*(1/cos^2x)/(-2/s

已知tan(x-y)=2,tan(y+π/5)=3,则tan(x+π/5)等于(x+π/5)=(x-y)+(y+π/5)tan(x+π/5)=[tan(x-y)+tan(y+π/5)]/[1-tan(x-y)*tan(y+π/5)]=(2+

tanπx/x+2的极限,x趋于-2

lim(2-x)^tanπx/2x趋近1这个是1^oo型的,运用重要的极限准则解题即可,具体如下：x→1时lim（2-x）^tan（πx）/2=x→1时lim[1+(1-x）]^1/(1-x)*(1-x)*tan（πx）/2=x→1时e^l

高数x趋向1x^(tan(πx/2))记y=x^(tan(πx/2)),则lny=tan(πx/2)*lnx=lnx/cot(πx/2),0/0型,则limlny=limlnx/cot(πx/2)=lim(1/x)/{-π/2[csc(πx

求极限：sinx^tanx(x→π/2)1

lim(x→1)(1-x)tan(π/2)xtan=sin/cos带入用罗比大法则上下求导再代入x=1即可lim(x→1)(1-x)=0lim(x→1)tan(π/2)x=正无穷lim(x→1)(1-x)tan(π/2)x=0tan(pi/