Question

高一数学题 求证：(1+cscα+cotα)/(1+cscα-cotα)=cscα+cotα

Answer 1

cscα=1/sina
cotα=cosa/sina
1+cscα+cotα= (sina+1+cosa)/sina
1+cscα-cotα=(sina+1-cosa)/sina
(1+cscα+cotα)/(1+cscα-cotα)= (sina+1+cosa)/(sina+1-cosa)
cscα+cotα =(1+cosa)/sina
(sina+1+cosa)* sina =sin²a+sina+sinacosa ①
(sina+1-cosa)*(1+cosa)=sina+sinacosa+1-cos²a =sina+sinacosa+sin²a ②
2个相等
(sina+1+cosa)* sina =(sina+1-cosa)*(1+cosa)
(sina+1+cosa)/(sina+1-cosa)=(1+cosa)/sina
∴ (1+cscα+cotα)/(1+cscα-cotα)=cscα+cotα