In this problem
the angle theta lies on the II quadrant
so
the tangent is negative
Remember that
[tex]1+\tan ^2(\theta)=\sec ^2(\theta)[/tex]
substitute the given value
[tex]1+\tan ^2(\theta)=(-\frac{\sqrt[]{143}}{11})^2[/tex]
[tex]\tan ^2(\theta)=\frac{143}{121}-1[/tex]
[tex]\tan ^2(\theta)=\frac{22}{121}[/tex]
[tex]\tan ^{}(\theta)=-\frac{\sqrt[]{22}}{11}[/tex]
