$\cos ec\theta +\cot \theta =\frac{1}{3}$

$\Rightarrow \frac{1+\cos \theta }{\sin \theta }=\frac{1}{3}$

$3+3\cos \theta =\sin \theta$

Squaring

$9{\cos }^2\theta +18\cos \theta +9=1-{\cos }^2\theta$

$\Rightarrow 10{\cos }^2\theta +18\cos \theta +8=0$

$\Rightarrow 5{\cos }^2\theta +9\cos \theta +4=0$

$\Rightarrow 5{\cos }^2\theta +5\cos \theta +4\cos \theta +4=0$

$\Rightarrow 5\cos \theta (\cos \theta +1)+4(\cos \theta +1)=0$

$\Rightarrow \cos \theta =-1, -\frac{4}{5}$

$\cos \theta =-\frac{4}{5}\to \theta$ lies in $2$nd or $4$th quadrant

But $\mathrm{cosec}\theta +\cot \theta$ is positive

So $\theta$ lies in $2$nd quadrant