Let [tex]a=\tan A[/tex] and [tex]b=\sin A[/tex]. Then
[tex]m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab[/tex]
[tex]\implies m^2-n^2=4\tan A\sin A[/tex]
and
[tex]mn=(a+b)(a-b)=a^2-b^2[/tex]
[tex]\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}[/tex]
The expression under the square root can be rewritten as
[tex]\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)[/tex]
Recall that
[tex]\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A[/tex]
so that
[tex]\tan^2A-\sin^2A=\sin^2A\tan^2A[/tex]
and assuming [tex]\sin A>0[/tex] and [tex]\tan A>0[/tex], we end up with
[tex]4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A[/tex]
so that
[tex]m^2-n^2=4\sqrt{mn}[/tex]
as required.
