Question

高三三角函数题,1、求证：(1+SecA+tanA)/(1+SecA-tanA)=(1+sinA)/cosA2、△ABC

Answer 1

1.(1+seca+tana)cosa=cosa+1+sina

(1+seca-tana)(1+sina)=1+1/cosa-tana+sina+sina/cosa-(sina)^2/cosa=1+cosa+sina

2.cosb=cos[a-(a-b)]=cosacos(a-b)+sinasin(a-b)其中cos(a-b)=3*10^(-1/2) sin(a-b)=-10^(-1/2)故得出答案是)(9/5)*10^(-1/2)