In a right triangle $ABC$,right-angled at $B$. If $\tan A = 1$,then verify that $2 \sin A \cos A = 1$.

(N/A) In $\triangle ABC$,$\tan A = \frac{BC}{AB} = 1$ (see figure).
i.e.,$BC = AB$.
Let $AB = BC = k$,where $k$ is a positive number.
Now,$AC = \sqrt{AB^2 + BC^2}$
$= \sqrt{k^2 + k^2} = \sqrt{2k^2} = k\sqrt{2}$.
Therefore,$\sin A = \frac{BC}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$ and $\cos A = \frac{AB}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}$.
So,$2 \sin A \cos A = 2 \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} \right) = 2 \left( \frac{1}{2} \right) = 1$,which is the required value.