tricky GCD Question I am trying to show that the $ gcd (\frac {n}{gcd(n,d)} , \frac {d}{gcd(n,d)} ) = 1 $. 
My steps let $gcd(n,d) = K = xn+yd$ then I need to find some linear combination of $\frac {n} {K} $ and $\frac {d} {K} $ that gives 1. Any hints would be appreciated.

We have
$$ x \cdot \frac nK + y \cdot \frac dK = \frac{xn + yd}K = 1. $$
Hence $\gcd(\frac nK, \frac dK) = 1$.
