If $\tan A+\sin A=m,\tan A-\sin A=n$ , then $m^2-n^2$

$m-n=2\sin A\Rightarrow \text{cosecA=}\frac{2}{m-n}$

${\text{cosec}}^{2}A-{\cot }^{2}A=1$

${\left(\frac{2}{m-n}\right)}^2-{\left(\frac{2}{m+n}\right)}^2=1$

$\frac{4\left({\left(m+n\right)}^2-{\left(m-n\right)}^2\right)}{{\left(m-n\right)}^2{\left(m+n\right)}^2}=1$

$4\left(4mn\right)={\left(m^2-n^2\right)}^2$

$\therefore m^2-n^2=4\sqrt{mn}$