两题三角函数求证明...证明两边相等.求助.1.sec(a)csc(a)+cot(a)/tan(a)-1=[tan(a)/1-cot(a)]-12.cot(a)+csc(a)=sin(a)/1-cos(a)能教教我吗?谢谢!最好有过程..

Question

两题三角函数求证明...证明两边相等.求助.
1. sec(a) csc(a) + cot (a) / tan(a)-1 =[tan (a)/ 1-cot (a)] -1
2.cot (a) + csc (a) = sin(a)/ 1-cos(a)
能教教我吗?谢谢!最好有过程..

Answer 1

1、sec(a) csc(a) + cot (a) / tan(a)-1 =[tan (a)/ 1-cot (a)] -1
<==>1/(sinacosa)+1/(tan²a-tana)=tan²a/(tan-1)-1
<==>1/(sinacosa)+cos²a/(sin²a-sinacosa)=sin²a/(sinacosa-cos²a)-1
<==>1/sinacosa=(sin²a-sinacosa+cos²a)/(sinacosa-cos²a)+cos²a/(sinacosa-sin²a)
<==>1/sinacosa=(1-sinacosa)/[cosa(sina-cosa)]-cos²a/[sina(sina-cosa)]
<==>1/sinacosa=(sina-sin²acosa-cos²acosa)/[sinacosa(sina-cosa)]
<==>sina-cosa=sina-sin²acosa-cos²acosa
<==>cosa=sin²acosa+cos²acosa
<==>1=sin²a+cos²a
以上各步可逆,证毕
2、cot (a) + csc (a) = sin(a)/ 1-cos(a)
<==>cosa/sina+1/sina=sina/(1-cosa)
<==>(cosa+1)/sina=sina/(1-cosa)
<==>sin²a=(1-cosa)(1+cosa)
<==>sin²a=1-cos²a
<==>sin²a=sin²a
以上各步可逆,证毕