∵tanα=2tan(α/2)/(1-tan²(α/2))
=(√5-1)/[1-(5-2√5+1)/4]
=2(√5-1)/(√5-1)=2
cos2α-sin2a=(1-tan²α)/(1+tan²α)-2tanα/(1+tan²α)
=(1-4)/(1+4)-4/(1+4)
=-3/5-4/5=-7/5