√(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
以上兩式 均沒考慮 符號(hào) 問題 ,即說明上式均可取 相反數(shù)
兩式相乘得 - 4 ,即也可能取到 4
因?yàn)樯鲜經(jīng)]討論,討論的話也簡(jiǎn)單,sinx ,cosx與0比較即可