Question

If $\tan A=\sqrt{2}-1$, prove that
$\frac{\tan A}{1+{\tan }^{2}A}=\frac{\sqrt{2}}{4}$

Answer 1

$\frac{\tan A}{1+{\tan }^{2}A}$

$=\frac{\sqrt{2}-1}{1+(\sqrt{2}-1{)}^2}$

$=\frac{\sqrt{2}-1}{1+2+1-2\sqrt{2}}$

$=\frac{\sqrt{2}-1}{2\sqrt{2}(\sqrt{2}-1)}$

$=\frac{\sqrt{2}}{4}$