Certainly! Let's analyze the given expression step by step.
We need to find what the expression [tex]\(\frac{2 \tan \left(10^{\circ}\right)}{1-\tan ^2\left(10^{\circ}\right)}\)[/tex] simplifies to in terms of a single trigonometric ratio.
Recall the double-angle identity for tangent:
[tex]\[
\tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)}
\][/tex]
If we let [tex]\( x = 10^\circ \)[/tex], we get:
[tex]\[
\tan(2 \times 10^\circ) = \frac{2 \tan(10^\circ)}{1 - \tan^2(10^\circ)}
\][/tex]
Simplifying this expression, we obtain:
[tex]\[
\tan(20^\circ) = \frac{2 \tan(10^\circ)}{1 - \tan^2(10^\circ)}
\][/tex]
Thus,
[tex]\[
\frac{2 \tan \left(10^{\circ}\right)}{1-\tan ^2\left(10^{\circ}\right)} = \tan(20^\circ)
\][/tex]
Therefore, the given expression [tex]\(\frac{2 \tan \left(10^{\circ}\right)}{1-\tan ^2\left(10^{\circ}\right)}\)[/tex] is best represented by [tex]\(\tan(20^\circ)\)[/tex].
So, the correct answer is:
[tex]\[
\boxed{\tan \left(20^{\circ}\right)}
\][/tex]
