已知sinx+sin(3/2π+x)=根号2,求tanx+tan(3/2π-x)

题目

已知sinx+sin(3/2π+x)=根号2,求tanx+tan(3/2π-x)

答案

tanx+tan(3/2π-x)
=tanx+ctgx
=sinx/cosx +cosx/sinx
=1/sinx*cosx
原式=sinx+sin(x-π/2)=根号2
(sinx-cosx)平方=2
1-2*sinx*cosx=2
sinx*cosx=-1/2
所以 tanx+tan(3/2π-x) =-2