结果一 题目 sina cosa tana cota seca csca分别是直角三角形的那个边比那个边 答案 sin是 对边/斜边cos是 邻边/斜边tan是 对边/邻边cot是 邻边/对边sec是 斜边/邻边csc是 斜边/对边相关推荐 1sina cosa tana cota seca csca分别是直角三角形的那个边比那个边 ...
答案 原式=(1/sina-sina)(1/cosa-cosa)(sina/cosa+cosa/sina) =[(1-sin²a)/sina][(1-cos²a)/cosa][(sin²a+cos²a)/sinacosa] =[cos²a/sina][sin²a/cosa](1/sinacosa) 约分 =1 相关推荐 1 (csca-sina)(seca-cosa)(tana+cota) 反馈 收藏 ...
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代数输入 三角输入 微积分输入 矩阵输入 sec(A)−sin(A) 求值 2cos(A)−sin(2A)+2 关于A 的微分 cos(A)tan(A)−cos(A)
(1+cotA-cosecA)(1+tanA+secA)=2 03:57 (sintheta+cosectheta)^(2)+(costheta+sectheta)^(2)=tan^(2)theta+cot^(2)... 02:08 tan^(2)theta+cot^(2)theta+2=sec^(2)thetacosec^(2)theta. 01:49 (cosA)/(1-tanA)-(sin^(2)A)/(cosA-sinA)=sinA+sinB. 02:51 (tantheta+sintheta)...
( (sinA-cscA) ) ( (cosA-secA) ) ( (tanA+cotA) ) $=\left ( {sinA-\dfrac {1} {sinA}} \right )\left ( {cosA-\dfrac {1} {cosA}} \right )\left ( {\dfrac {sinA} {cosA}+\dfrac {cosA} {sinA}} \right )$ $=\dfrac {{sin}^{2}A-1} {sinA}\cdot \dfrac {{...
(tana-cota)/(seca-csca)=sina+cosa 证明 相关知识点: 试题来源: 解析 证明:tana-cota=sina/cosa-cosa/sina=(sin²a-cos²a)/(cosasina)seca-csca=1/cosa-1/sina=(sina-cosa)/(sinacosa)(tana-cota)/(seca-csca)=sina+cosa反馈 收藏 ...
解：cosa=-5/13，则：sina=-12/13或12/13 tana=sina/cosa=±12/5 cota=cosa/sina=±5/12 seca+csca=-13/5 +(-13/12)=-221/60 或seca+csca=-13/5 +(13/12)=-91/60 不理解之处欢迎追问，满意敬请采纳...sina
用三角函数的定义就可以了.即sinA＝x/r tanA=y/x cotA=x/y secA=r/x cscA=r/y 两边同时代入即可证明.
Prove the following trigonometric numbers. (1+cotA+tanA)(sinA-cosA)=(secA)/(cosec^(2)A)-(cosecA)/(sec^(2)A) View Solution Prove the following trigonometric numbers. (tan^(3)A)/(1+tan^(2)A)+(cot^(3)A)/(1+cot^(2)A)=secA*cosecA-2sinA*cosA ...