Since [tex]\cot(x)=\frac{2}{3}[/tex] and [tex]\cot^{2} x+1=\csc^{2} x[/tex], we know that:
[tex]\left(\frac{2}{3} \right)^{2}+1=\csc^{2} x\\\\\frac{13}{9}=\csc^{2} x\\\\\csc x=\frac{\sqrt{13}}{3}[/tex]
If [tex]\csc x=\frac{\sqrt{13}}{3}[/tex], this means that [tex]\sin x=\frac{3}{\sqrt{13}}[/tex] and by the Pythagorean identity,
[tex]\sin^{2} x+\cos^{2} x=1\\\left(\frac{3}{\sqrt{13}} \right)^{2}+\cos^{2} x=1\\\frac{9}{13}+\cos^{2} x=1\\\cos^{2} x=\frac{4}13}\\\cos x=\frac{2}{\sqrt{13}}[/tex]
- Using the double angle formula for sine, [tex]\sin(2x)=2\left(\frac{3}{\sqrt{13}} \right)\left(\frac{2}{\sqrt{13}} \right)=\boxed{\frac{12}{13}}[/tex]
- Using the double angle formula for cosine, [tex]\cos(2x)=1-2\left(\frac{3}{\sqrt{13}} \right)^{2}=\boxed{-\frac{5}{13}}[/tex]
- So, since tan=sin/cos, [tex]\tan (2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\frac{12}{13}}{-\frac{5}{13}}=\boxed{-\frac{12}{5}}[/tex]
