If sinθ = 2/3 and θ is located in Quadrant II, then tan2θ = _____.

Answer:[tex] -4 \sqrt{5} [/tex]Step-by-step explanation:Quadrant 2 means cosine is negative.So [tex] \sin(\theta)=\frac{2}{3} =\frac{\text{ opp }}{\text{ hyp }} [/tex]So the adjacent side is [tex] \sqrt{3^2-2^2}=\sqrt{9-4}=\sqrt{5} [/tex]So [tex] \cos(\theta)=-\frac{\sqrt{5}}{3} [/tex]Now to find [tex] \tan(2 \theta) [/tex][tex] \tan(2 \theta) =\frac{2\tan(\theta)}{1-\tan^2(\theta)}[/tex]We will need [tex] \tan(\theta) [/tex] before proceeding. [tex] \tan(\theta) =\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{2}{3}}{\frac{-\sqrt{5}}{3}}=\frac{-2}{\sqrt{5} } [/tex]Now plug it in and the rest is algebra.[tex] \tan(2 \theta) =\frac{2\tan(\theta)}{1-\tan^2(\theta)} =\frac{2 (\frac{-2}{\sqrt{5}}}{1-\frac{4}{5}} [/tex]Now the algebra, the simplifying.... We need to get rid of the compound fraction.  We will multiply top and bottom by [tex] 5 \sqrt{5} [/tex]This will give us [tex] \frac{-4(5)}{5 \sqrt{5}-4 \sqrt{5}} [/tex][tex] \frac{-20}{\sqrt{5}} [/tex]Multiply top and bottom by [tex] \sqrt{5} [/tex][tex] \frac{-20 \sqrt{5}}{5} [/tex]The answer reduces to [tex] -4 \sqrt{5} [/tex]