Question

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

Answer 1

$\tan A=\sqrt{2}-1$

$\tan A=\frac{BC}{AB}=\frac{\sqrt{2-1}}{1}$

$A{C}^2=B{A}^2+B{C}^2=1^2+(\sqrt{2}-1{)}^2=1+2-2\sqrt{2}+1$

$AC=\sqrt{4-2\sqrt{2}}$

$\sin A=\frac{BC}{AC}=\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}\cos A=\frac{AB}{AC}=\frac{1}{\sqrt{4-2\sqrt{2}}}$

$\sin A\times \cos A=\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}\times \frac{1}{\sqrt{4-2\sqrt{2}}}=\frac{\sqrt{2}-1}{4-2\sqrt{2}}$

$=\frac{\sqrt{2}-1}{2\sqrt{2}(\sqrt{2}-1)}=\frac{1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}$

$\cos A\times \frac{\sin A\times cosA}{\cos A}=\frac{\sqrt{2}}{4}$

Multiply numerator & denominator by cosA

$\frac{\tan A}{{\sec }^{2}A}=\frac{\sqrt{2}}{4}\Rightarrow \frac{\tan A}{1+{\tan }^{2}A}=\frac{\sqrt{2}}{4}$