设 x=arccos(1677/1685)
∴ 0<x<π/2，cosx=1677/1685
∴ sinx=√(1-cos²x)=164/1685
原式=cscx=1/sinx=1685/164
回答2：
csc[arccos(1677/1685)]
设x=arccos(1677/1685)
则cosx=1677/1685
那么，csc[arccos(1677/1685)]
=cscx
=1/sinx
=1/√(1-cos²x)
=1/√[1-(1677/1685)²]
=1/√(26896/1685²)
=1/√( 164 ²/1685²)
=1/( 164 /1685)
=1685/164
所以csc[arccos(1677/1685)]=1685/164
