Question

化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a)

答案 过程

Answer 1

原式可化为：根号1-c0sa/1+cosa+根号1-cosa/1+cosa
=2倍根号1-ccosa/1+cosa
=2倍根号1-cos平方a/括号1+cosa括号的平方
=2sina/1+cosa
解答完毕

Answer 2

化简：√{[1-cos(2π+α)]/[1+os(2π+α)]}+√{[1+cos(2π+α)]/[1-cos(2π-α)]}.
原式=√[(1-cosα)/(1+cosα)]+√[(1+cosα)/(1-cosα)].
=√[(1-cosα)(1+cosα)/(1+cosα)^2]+√[(1+cosα)(1-cosα)/(1-cosα)^2].
=√(1-cos^2α)/(1+cosα)+√(1-cos^2α)/(1-cosα).
=|sinα|/(1+cosαα)+|sinα|/(1-cosα).
=|sinα|[(1-cosα)+(1+cosα)]/(1-cos^2α).
=|sinα|*2/|sinα|.
∴原式=2 .