Question

If $\tan A+\sin A=m$ and $\tan A-\sin A=n$, then
show that $m^2-n^2=4\sqrt{mn}$.

Answer 1

$m^2-n^2=(m+n)(m-n)$
$=2\tan A.2\sin A=4\tan A\sin A$,
$4\surd \left(mn\right)=4\surd {(\tan }^2-A{\sin }^{2}A)$
$=4\sin A\surd {(\sec }^{2}A-1)=4\sin A\tan A$.
$\therefore m^2-n^2=4\surd \left(mn\right)$.