Pure Mathematics - Inequality

Let a1=a2=...=am=sin^2θ/m ; am+1=am+2=...=am+n=cos^2θ/n
a1+a2+...+am+n=1
Apply AM>=GM
[(a1+a2+...+am+n)/m+n]^(m+n)>=a1a2...am+n
[1/(m+n)]^(m+n)>=(sin^2mθcos^2nθ)/(m^m)(n^n)
sin^2mθcos^2nθ<=(m^m)(n^n)/[(m+n)]^(m+n)
when the equality holds
sin^2θ/m=cos^2θ/n
tan^2θ=m/n
or tanθ=sqrt(m/n)