Question

数学题：帮忙化简下吧

已知sin a + cos a = -（√2+1）/ 2； sin a cos a=m/2
求 cos a/ 1-cot^2 a + sin a/ 1-tan^2 a
速度啊！很简单的哦
问题已解决了哦！

Answer 1

cos a/ 1-cot^2 a + sin a/ 1-tan^2 a
=cos a/ [1-(csc^2 a-1)] + sin a/ [1-(sec^2 a-1)]
=cos a/ (2-csc^2 a) +sin a/(2-sec^2a)
=cos a/ (2-1/sin^2 a) +sin a/(2-1/cos^2a)
=cos a/ (2-1/sin^2 a) +sin a/(2-1/cos^2a)
=cos a/ [(2sin^2 a-1)/sin^2 a] +sin a/[(2cos^2-1)/cos^2a]
=cos asin^2 a/ (2sin^2 a-1) +sin acos^2a/(2cos^2a-1)
=cos asin a[sina/ (2sin^2 a-1)+cosa/(2cos^2a-1)]
=m/2*[sina/ (2sin^2 a-1)+cosa/(2cos^2a-1)]
=m/2*[sina(2cos^2-1)/(2sin^2 a-1)+cosa(2sin^2 a-1)/(2cos^2a-1)]
=m/2*[(2sinacos^2a-sina+2cosasin^2 a-cosa)/(2sin^2 a-1)(2cos^2-1)]
=m/2*[(2sinacos^2a+2cosasin^2 a-sina-cosa)/(2sin^2 a-1)(2cos^2-1)]
=m/2*[(mcosa+msina-sina-cosa)/(2sin^2 a-1)(2cos^2a-1)]
=m/2*[(m-1)(sina+cosa)/(2sin^2 a-1)(2cos^2a-1)]
=m/2*[(m-1)(sina+cosa)/[4sin^2cos^2 a-2(2sin^2 a+cos^2a)+1]
=m/2*[(m-1)(sina+cosa)/[m^2-2+1]
=m/2*[(m-1)(sina+cosa)/[m^2-1]
=m/2*[(m-1)(sina+cosa)/[(m-1)(m+1)]
=m/2*(sina+cosa)/(m+1)
=m(sina+cosa)/2(m+1)
= -（√2+1）/ 2*m/2(m+1)
= -(√2+1)m/(4m+4)

Answer 2

sin a/ 1 这个除以1什么意思？