Question

If $\tan A+\sin A=m$ and $\tan A-\sin A=n$. Prove that $(m^2-n^2{)}^2=16mn$.

Answer 1

Given that,

$\tan A+\sin A=m$

$\tan A-\sin A=n$

$mn=(\tan A+\sin A)(\tan A-\sin A)={\tan }^{2}A-{\sin }^{2}A$

$={\sin }^{2}A(\frac{1}{{\cos }^{2}A}-1)={\sin }^{2}A({\sec }^{2}A-1)$

$={\sin }^{2}A{\tan }^{2}A$

$(m^2-n^2{)}^2=[(\tan A+\sin A{)}^2-(\tan A-\sin A{)}^2{]}^2$

$=(4\tan A\sin A{)}^2$

$=16{\tan }^{2}A{\sin }^{2}A$

$=16mn$

$\therefore (m^2-n^2{)}^2=16mn$

Hence proved.