From the question it is given that,

$m=asecC+btanC$

$n=atanC+bsecC$

We have to prove that, ${{m}^{2}}-{{n}^{2}}={{a}^{2}}-{{b}^{2}}$

Then,

${{m}^{2}}-{{n}^{2}}={{\left( a\sec C+b\tan C \right)}^{2}}-{{\left( a\tan C+b\sec C \right)}^{2}}$

We know that,${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

${{a}^{2}}{{\sec }^{2}}C+{{b}^{2}}{{\tan }^{2}}C+2a\sec Cb\tan C-\left( {{a}^{2}}{{\tan }^{2}}C+{{b}^{2}}{{\sec }^{2}}C+2ab\sec C\tan C \right)$

$={{\sec }^{2}}C\left( {{a}^{2}}-{{b}^{2}} \right)+{{\tan }^{2}}C\left( {{b}^{2}}-{{a}^{2}} \right)$

$=\left( {{a}^{2}}-{{b}^{2}} \right)\left[ {{\sec }^{2}}C-{{\tan }^{2}}C \right]$

Also we know that, ${{\sec }^{2}}C-{{\tan }^{2}}C=1$
$=\left( {{a}^{2}}-{{b}^{2}} \right)$

Hence it is proved that, ${{m}^{2}}-{{n}^{2}}={{a}^{2}}-{{b}^{2}}$