數學3角比一問(20分)

證明{[sin(90˚-n)sin n]/[(sin n)-1]}+{[cos(90˚-n)cos n]/[(sin n)+1]}≣-(2sin n)/tan(90˚-n)。
{[sin(90˚-n)sin n]/[(sin n)-1]}+{[cos(90˚-n)cos n]/[(sin n)+1]}≣-(2sin n)/tan(90˚-n)
L.H.S. (left hand side)= {[sin(90˚-n)sin n]/[(sin n)-1]}+{[cos(90˚-n)cos n]/[(sin n)+1]}
=(cos n)(sin n)/[(sin n)-1]+(sin n)(cos n)/[(sin n)+1]
=[(sin n)+1](cos n)(sin n)+[(sin n)-1](sin n)(cos n)/{[(sin n)+1][(sin n)-1]}
=(sin n)(cos n){[(sin n)+1]+[(sin n)-1]} /(sin²n-1)
=-(sin n)(cos n)(2sin n)/(1-sin²n)
=-(2sin²n)(cos n)/cos²n
=-(2sin n)(sin n)/cos n
=-(2sin n)(sin n/cos n)
=-(2sin n)(tan n)
R.H.S.(right hand side)=-(2sin n)/tan(90˚-n)
=-(2sin n)(tan n)
∵L.H.S.=R.H.S.
∴這是一條恆等式。