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

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In $\triangle ABC , \tan A =\frac{ BC }{ AB }=1$ (see $Fig.$)

i.e. $BC = AB$

Let $AB = BC =k,$ where $k$ is a positive number.

Now,$AC=\sqrt{ AB ^{2}+ BC ^{2}}$

$=\sqrt{(k)^{2}+(k)^{2}}=k \sqrt{2}$

Therfore, $\sin A=\frac{ BC }{ AC }=\frac{1}{\sqrt{2}} \quad$ and $\cos A =\frac{ AB }{ AC }=\frac{1}{\sqrt{2}}$

So, $\quad 2 \sin A \cos A =2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=1,$ which is the required value.

1043-s4

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