(1 3tan^2x)^cotx

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

(1+cotx)sin^2x=sin^2x+sinxcosx2sin(x+π/4)sin(x-π/4)=根号2(sinx+cosx)*根号2/2(sinx-cosx)=sin^2x-cos^2xf(x

原式=(sinx/cosx+cosx/sinx)cos²x=(sin²x+cos²x)*cos²x/(sinxcosx=1*cos²x/sinxcos

y=tanπx周期T=π/W=π/π=1y=cotx-tanx=cosx/sinx-sinx/cosx=(cos^2x-sin^2x)/sinxcosx=2cos2x/sin2x=2cot2x周期T=

cotx+coty=301/tanx+1/tany=30(tanx+tany)/tanxtany=3025/tanxtany=30tanxtany=25/30=5/6,tan(x+y)=(tanx+t

其实可以这样解:原极限=lim(x->0)[(tanx)^2-x^2]/[x^2*(tanx)^2]=lim(x->0)[(tanx)^2-x^2]/x^4=lim(x->0)[(tanx+x)/x]

=∫arccotxdx∧2=x∧2arccotx+∫x∧2/（1＋x∧2）dx=x∧2arccotx+∫（1+1/x∧2）dx=x∧2arccotx+x-1/x+c

详解如下：

cotx+coty=1/tanx+1/tany=(tanx+tany)/(tnaxtany)即30=25/(tanxtany),所以tanxtnay=5/6tan(x+y)=(tanx+tany)/(

(tanx+cotx)cos^2x=(sinx/cosx+cosx/sinx)cos²x=(sin²x+cos²x)*cos²x/(sinxcosx）=1*co

f(x)=sin(k兀-x)/sinx-cosx/cos(k兀-x)+tan(k兀-x)/tanx-cotx/cot(k兀-x)=sin(k兀-x)/sinx-cosx/cos(k兀-x)-1+1当k

=lim(1/x)^2-(1/tanx)^2=lim(x^2-tan^2x)/(x^2·tan^2x)=lim(x^2-tan^2x)/(x^4)【等价无穷小代换】=lim(2x-2tanx/cos&

lim(x→0)cotx[2x/(1-x)]=lim(x→0)2x/[tanx(1-x)]x→0tanx与x价=lim(x→0)2x/[x(1-x)]=lim(x→0)2/(1-x)=2

(sinx)^2tanx=[1-(cosx)^2]tanx=tanx-(cosx)^2tanx=tanx-(cosx)^2*sinx/cosx=tanx-sinxcosx(cosx)^2cotx=[1

再问：可以具体点吗再问：数学白痴一个再答：用的是无穷小等价代换。不需要什么过程。

你的结果和答案是一样的cscx-cotx=1/sinx-cosx/sinx=(1-cosx)/sinx=2sin²(x/2)/2sin(x/2)cos(x/2)=sin(x/2)/cos(x

1.[㏑(x-π/2)]/tanx当x趋于π/2时的极限=lim(x->π/2)1/(x-π/2)/sec²x=lim(x->π/2)cos²x/(x-π/2)=lim(x->π/

lim(x→0)[tan(π/4-x)]^(cotx)=lim(x→0){e^[cotx*ln(tan(π/4-x))]}只需要求lim(x→0)[cotx*ln(tan(π/4-x))]；lim(x