What is [tex]\frac{2 \tan \left(10^{\circ}\right)}{1-\tan ^2\left(10^{\circ}\right)}[/tex] expressed as a single trigonometric ratio?

A. [tex]-\tan \left(20^{\circ}\right)[/tex]
B. [tex]-\tan \left(5^{\circ}\right)[/tex]
C. [tex]\tan \left(5^{\circ}\right)[/tex]
D. [tex]\tan \left(20^{\circ}\right)[/tex]



Answer :

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]