√(1-sinx)/(1+sinx)-√(1+sinx)/(1-sinx)
=( sin(x/2) - cos(x/2) ) / ( sin(x/2) + cos(x/2) ) - ( sin(x/2) + cos(x/2) )/ ( sin(x/2) - cos(x/2) )
= [ ( sin(x/2) - cos(x/2) )^2 - ( sin(x/2) + cos(x/2) )^2 ] / ( sin(x/2)^2 - cos(x/2)^2 )
= 2* sinx/cosx
√(1-cosx)/(1+cosx)-√(1+cosx)/(1-cosx)
= sin(x/2)/cos(x/2) - cos(x/2)/sin(x/2)
= ( sin(x/2)^2 - cos(x/2)^2 ) / ( sin(x/2) * cos(x/2) )
= - cosx / (1/2 * sinx)
= - 2 * cosx/sinx
以上两式 均没考虑 符号 问题 ,即说明上式均可取 相反数
两式相乘得 - 4 ,即也可能取到 4
因为上式没讨论,讨论的话也简单,sinx ,cosx与0比较即可