求证tan^2x+1/tan^2x=[2(3+cos4x)]/(1-cos4x)

来源：学生作业学帮网编辑：学帮网时间：2024/05/18 04:27:33

左边=sin^2x/cos^2 + cos^2/sin^2x=(sin^4x+cos^4x)/(cos^2xsin^2x)=(sin^2x+cos^2x)^2-2sin^2xcos^2x/(sin^2xcos^2x)=(1-sin^2(2x)/2)/(sin^2(2x)/4)=[4-2(1-cos4x)/2]/[(1-cos4x)/2]=右边

求证tan(x+y)*tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^2y 1 求证：tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)2 已知a+b+c=npai(n属于Z),求证：tan(a)+tan(b)+tan(c)=tan(a)tan(b)tan(c)（提示：在等式a+b=npai-b同时取正切）求证：tan x/2=sin x/(1+cos x) 求证：1-sin(x)=cos(x)*tan(x/2) 1-tan*2x/1+tan*2x=cos*2x-sinx求证求证tan^2x+1/tan^2x=[2(3+cos4x)]/(1-cos4x) 求证：1-2sinxcosx/cos方x-sin方x=1-tan/+tan 求证：(tan的平方x)+(1/tan的平方x)=2*（3+cos4x)/(1-cos4x) 求证：tan的平方x+1/tan的平方x=2(3+cos4X)/1-COS4x 求证：tan平方x+1/tan平方x=2(3+cos4x)/1-cos4x 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+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x 求证tan^x-sin^2x=tan^2x*sin^2x 求证；1-2sin Xcos X/cos2 X-sin2 X=1-tan X/1+tan X tan^2x+1/tan^2=2(3+cos4x)/1-cos4x求证求证1+sinx/cosx=tan(x/2), 求证（1—tan^2X)/(1+tan^2X)=cos^2X—sin^2X 求证：tan(x/2+派/4)+tan(x/2-派/4)=2tanx{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到这道题为什么第二步的+到了第三步变成了-1.tan(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+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-