In the right angle triangle $\;ABC$ , the right angle at $B$ , the ratio of $\;AB$ to $\;AC.$ is $1:\sqrt 2$ . Find the value of $\dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$ .

Hint: At first, we will draw the triangle $\;ABC$ . The ratio of $\;AB$ to $\;AC$ is given $1:\sqrt 2$ . We will assume a constant $k$ and get the length of two sides. Then I will find the value of $\tan A$ . Then we will put the value of $\tan A$ in $\dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$ . In this way, we will get the value.

Complete step-by-step solution:
We have given a right angle triangle $\;ABC$ . In the triangle, the right angle is at $B$ . We need the value of $\tan A$ . So we will draw $\;AB$ as the base.
So the triangle is;

Now, this is given that the ratio of $\;AB$ to $\;AC$ is $1:\sqrt 2$ .
Let us choose an arbitrary constant $k$ such that;
$AB = k$
And $AC = \sqrt 2 k$
$\;ABC$ is a right angle triangle where $\;AB$ is the base and $\;AC$ is the hypotenuse. So, another side will be $BC = \sqrt {A{C^2} - A{B^2}}$ .
Putting the values we get;
$\Rightarrow BC = \sqrt {{{\left( {\sqrt 2 k} \right)}^2} - {k^2}}$
Simplifying we get;
$\Rightarrow BC = \sqrt {2{k^2} - {k^2}}$
Simplifying we get;
$BC = k$ .
As we know that $\tan \theta = \dfrac{{base}}{{perpendicular}}$ .
Hence $\tan A = \dfrac{{AB}}{{BC}}$ .
Then $\tan A = \dfrac{k}{k}$ .
Simplifying we get;
$\tan A = 1$ .
Now we will put the value of $\tan A$ in $\dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$ and we will get;
$\dfrac{{2\tan A}}{{1 + {{\tan }^2}A}} = \dfrac{2}{{1 + 1}}$
Simplifying the above equation we get;
$\Rightarrow \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}} = \dfrac{2}{2}$
Simplifying the above equation we get;
$\Rightarrow \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}} = 1$ .

So the value of $\dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$ is $1$ .

Note: For this kind of problem at first, we have to draw the triangle otherwise we couldn’t find the value of the trigonometric functions. Here the ratio of the two sides is given. By using that this is easy to find out another side. We know that if the three sides of a right-angle triangle are known then we will get all the trigonometric functions.