谁能帮我解决一道数学题???

搞定，就烦了点

先把(tanx)^2=sinx^2/cos^2
=(1-cos2x)/(1+cos2x)
=-1+2/(1+cos2x)
然后算y=1/(1+cos2π/7)+1/(1+cos4π/7)+1/(1+cos6π/7)
即可
通过通分和积化和差，可以算得
y=(3+3t)/(9t/4+5/4)
t=cos2π/7+cos4π/7+cos6π/7
=(2sin2π/7)*(cos2π/7+cos4π/7+cos6π/7)/(2sin2π/7)
=(sin12π/7)/(2sin2π/7)
=-1/2
带入即可，得y=12
z=(tanπ/7)^2+(tan2π/7)^2+(tan3π/7)^2
=-3+2y
=21

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