Question 6If $0\lt \theta\lt \pi/4$ is such that $\operatorname{cosec}\theta-\sec\theta=\sqrt{13}/6$, then $\cot\theta-\tan\theta$ equals$\frac{\sqrt{13}}6$$\frac{\sqrt{12}}6$$\frac{\sqrt5}6$$\frac{13}6$$\frac56$