In quadrant 2, [tex]\sin\theta>0[/tex] and [tex]\cos\theta<0[/tex]. Use the Pythagorean identity to establish that
[tex]\sin^2\theta+\cos^2\theta=1\implies\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac5{13}[/tex]
Then
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac{12}{13}}{-\frac5{13}}=-\dfrac{12}5[/tex]
