Respuesta :

We can use the following trigonometric identity to solve the exercise:

[tex]\frac{2\tan\theta}{1-\tan^2\theta}=\tan 2\theta[/tex]

In this case, θ = 15°. Then, we have:

[tex]\begin{gathered} \frac{2\tan15\degree}{1-\tan^215\degree}=\tan 2(15\degree) \\ \frac{2\tan15\degree}{1-\tan^215\degree}=\tan 30\degree \end{gathered}[/tex]

We can use a notable angles table to find the value of tan 30°. Finally, we have:

[tex]\frac{2\tan15\degree}{1-\tan^215\degree}=\frac{\sqrt[]{3}}{3}[/tex]

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