Question

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

Answer 1

In $△ABC$, $\angle ABC={90}^o$ $\therefore \tan A=\frac{BC}{AB}$
Since $\tan A=1$ (Given) $\frac{BC}{AB}=1$ $\therefore BC=AB$
Let $AB=BC=k$, where $k$ is a positive number.
Now, ${AC}^2={AB}^2+{BC}^2$ $\therefore AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{k^2+k^2}$
$\therefore AC=k\sqrt{2}$
$\therefore \sin A=\frac{BC}{AC}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}},\cos A=\frac{AB}{AC}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}$
$2\sin A\cos A=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=1$
$\therefore 2\sin A\cos A=1$