For this problem, we are given a certain trigonometric expression shown below:
[tex]\frac{2\tan(\frac{2\pi}{3})_{}}{1-\tan^2(\frac{2\pi}{3})}[/tex]And we need to represent it as a sin, tan, or cos of a double angle. The first thing we can do to make this easier is to assign a variable to 2pi/3, which we can see below:
[tex]x=\frac{2\pi}{3}[/tex]Now we can replace the expression above with the original one:
[tex]\frac{2\tan(x)_{}}{1-\tan^2(x)}[/tex]We know that a tangent is a division between a sine and a cosine. We can replace the tangent on the expression above with that division:
[tex]\frac{2\frac{\sin(x)}{\cos(x)}_{}}{1-\frac{\sin^2(x)}{\cos^2(x)}}[/tex]Now we need to find the LCM on the denominator, so we can transform the denominator into a single fraction.
[tex]\frac{2\frac{\sin(x)}{\cos(x)}_{}}{\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}}[/tex]Now, we have a division between two fractions. We need to conserve the first fraction and multiply it by the inverse of the second one.
[tex]\begin{gathered} 2\frac{\sin(x)}{\cos(x)}\cdot\frac{\cos^2(x)}{\cos^2(x)-\sin^2(x)} \\ \frac{2\sin(x)\cos^{}(x)}{\cos^2(x)-\sin^2(x)} \end{gathered}[/tex]We have a trigonometric identity that says the following:
[tex]2\sin (x)\cos (x)=\sin (2x)[/tex]And another that states this:
[tex]\cos ^2(x)-\sin ^2(x)=\cos (2x)[/tex]If we replace the two identities above on our fraction, we will obtain:
[tex]\frac{2\sin(x)\cos^{}(x)}{\cos^2(x)-\sin^2(x)}=\frac{\sin(2x)}{\cos(2x)}[/tex]The division between the sine and cosine is equal to the tangent, therefore we have:
[tex]\frac{\sin(2x)}{\cos(2x)}=\tan (2x)[/tex]We know the value for x, therefore we can replace it on the expression above and find the value for the tangent.
[tex]\begin{gathered} \tan (2\cdot\frac{2\pi}{3}_{}) \\ \tan (\frac{4\pi}{3}_{}) \end{gathered}[/tex]This concludes that:
[tex]\frac{2\tan(\frac{2\pi}{3})_{}}{1-\tan^2(\frac{2\pi}{3})}=\tan (\frac{4\pi}{3}_{})[/tex]The correct option is B, and the value we need to input on the blank box is 4pi/3.
