(1 3tan^2x)^cot^2x

x1x2=1x1=2+√3所以x2=2-√3则tanθ+cotθ=x1+x2=4tanθ+cotθ=sinθ/cosθ+cosθ/sinθ=4(sin²θ+cos²θ)/sinθc

sin^2(x)+tan^2(x)+csc^2(x)+sec^2(x)+cos^2(x)+cot^2(x)=31得cos^2(x)+sin^2(x)+tan^2(x)+csc^2(x)+sec^2(x

度数030456090180270sin01/2√2/2√3/210-1cos1√3/2√2/21/20-10tan0√3/31√3--0--cot--√31√3/30--0

我做的答案是2/sina因为tan(a/2)=sin(a/2)/cos(a/2)cot(a/2)=cos(a/2)/sin(a/2)所以原式=sin(a/2)/cos(a/2)+cos(a/2)/si

sin(3丌-a)/tan(a-5丌)xtan(-a+2丌)/cot(丌-a)xcos(4丌-a)/sin(-a-2丌)=sin(丌-a)/tan(a-丌)xtan(-a)/cot(丌-a)xcos(

sin(π-x)+sin(π+x)-cos(-x)+cos(2π-x)-tan(π+x)cot(π-x)=-sin(-x)-sin（x）-cos(x)+cos(-x)-[-tan(x)][-cot(-

tanx/(1-cotx)+cotx/(1-tanx)=tanx/(1-cosx/sinx)+cotx/(1-sinx/cosx)=sinxtanx/(sinx-cosx)+cosxcotx/(cos

tan^2x+cot^2x-(sin^2x*tan^2x+cos^2x*cot^2x)=tan^2x+cot^2x-sin^2x*tan^2x-cos^2x*cot^2x=tan^2x(1-sin^2

sin(2派-x)tan(派+x)cot(-x-派)/tan(3派-x)cos(派-x)=-sinxtanxcot(-x)/tan(-x)(-cosx)=sinxtanxcotx/tanxcosx=s

ezplot('tan(x)');gridon;ezplot('cot(x)');gridon;

1/(tanα+cotα)=sinαcosα/[(tanα+cotα)*sinαcosα]=sinαcosα/（sinα^2+cosα^2）=sinαcosα第二条,等号何在……

不确定你的问题是对每一个fun(x)进行定义域和值域求解还是其连乘之后.因此不能随便作答.而且这些在书上应该很容易找的,百度上也有啊?!祝顺利.

/>lim[ln(1-x)+tanπx/2]/cotπx=limln(1-x)/cotπx+limtanπx/2/cotπx=limsin²πx/[π(1-x)]+lim2tan²

左边=(sin²x/cos²x-cos²x/sin²x)/(sin²x-cos²x)=[(sin^4x-cos^4x)/cos²x

左边将tan(a)和cot(a)分别化成sin(a)和cos(a)直接利用公式1+tan^2(a)=1/cos^2(a)和sin^2(a)+cos^2(a)=1就可以证明,详细过程你最好自己来证.

[1-tan^2(a/2)]/tan(a/2)*[1+2tan^2(a/2)]/[1-tan^2(a/2)]=[1-tan^2(a/2)+2tan^2(a/2)]/tan(a/2)=[1+tan^2(

(1)(tanα)^3-(cotα)^3=2[(tana)²+(cota)²+1]（立方差公式）=2[(tana-cota)²+3]=2×（4+3）=14.(2)(sin

sin^2x+cos^2x=1所以左边=tan^2x-cot^2x=sin^2x/cos^2x-cos^2x/sin^2x=(sin^4x-cos^4x)/sin^2xcos^2x=(sin^2x+c

tan^2x+cot^2x=tan^2x+1/tan^2x=[(sinx)^4+(cosx)^4]/(sinxcosx)^2={[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^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^