[1/tan a/2-tan a/2]*(1+tan a*tan a/2)化简
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[1/tan a/2-tan a/2]*(1+tan a*tan a/2)化简
![[1/tan a/2-tan a/2]*(1+tan a*tan a/2)化简](/uploads/image/z/18587459-11-9.jpg?t=%5B1%2Ftan+a%2F2-tan+a%2F2%5D%2A%281%2Btan+a%2Atan+a%2F2%29%E5%8C%96%E7%AE%80)
[1/tan(a/2)-tan(a/2)]×](1+tan a×tan(a/2)]
={[cos²(a/2)-sin²(a/2)]/[cos(a/2)×sina/2]}×{[cosa×cos(a/2)+sina×sin(a/2)]/[cosa×cos(a/2)]}
={2[1/2+cosa/2-1/2+cosa]/[2cos(a/2)×sina/2]}×{cos(a-a/2)/[cosa×cos(a/2)]}
={2cosa/sina}×{cos(a/2)/[cosa×cos(a/2)]}
={2cosa/sina}×1/cosa
=2/sina
={[cos²(a/2)-sin²(a/2)]/[cos(a/2)×sina/2]}×{[cosa×cos(a/2)+sina×sin(a/2)]/[cosa×cos(a/2)]}
={2[1/2+cosa/2-1/2+cosa]/[2cos(a/2)×sina/2]}×{cos(a-a/2)/[cosa×cos(a/2)]}
={2cosa/sina}×{cos(a/2)/[cosa×cos(a/2)]}
={2cosa/sina}×1/cosa
=2/sina
2(tan a+tan b)
化简 (cot a/2 -tan a/2)(1+ tan a * tan a/2)
证明:(1+tan a+cot a)/(1+tan^2 a+tan a)-cot a/(1+ tan^2 a)=sin
tan A:tan B:tan C=1:2:3 求A:B:C
(cot (a/2) -tan (a/2))^2(1-tan a / tan (2a))
为什么 tan a +1/tan a=sin^2 a +cos^2 a/(sina cos a)
tan(a/2)-1/tan(a/2)=-2/tana 化简
已知tan(A-B/2)=2,tan(B-A/2)=-3 求:(1) tan[(A+B)/2] (2) tan(A+B)
tan(2兀-a)?等于
tan(B/2)tan(C/2)+tan(A/2)tan(B/2)+tan(A/2)tan(C/2)= A+B+C=18
tan(a+b)=3/4,tan(a-π/4)=1/2,那么tan(b+π/4)等于多少
已知A、B是锐角,求证(tan(π+A)+tan(-B))/(1/tan(3π-A)+tan(π/2-B))=tanA*