6. If $0<\theta<\frac{\pi}{4}$ is such that $\operatorname{cosec} \theta-\sec \theta=\frac{\sqrt{13}}{6}$, then $\cot \theta-\tan \theta$ equals
(A) $\frac{\sqrt{13}}{6}$;
(B) $\frac{\sqrt{12}}{6}$
(C) $\frac{\sqrt{5}}{6}$ :
(D) $\frac{13}{6}$;
(E) $\frac{5}{6}$.
6. Answer: (E).

Let $k=\sin \theta \cos \theta$. Then
$$
\frac{13}{36}=\left(\frac{1}{\sin \theta}-\frac{1}{\cos \theta}\right)^{2}=\frac{1-2 \sin \theta \cos \theta}{(\sin \theta \cos \theta)^{2}}=\frac{1-2 k}{k^{2}} .
$$

Solve the equation: $k=6 / 13(k=-6$ is rejected $)$. Then $\sin 29=2 k=12 / 13$ and
$$
\cot \theta-\tan \theta=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\sin \theta \cos \theta}=\frac{\cos 2 \theta}{\frac{1}{2} \sin 2 \theta}=\frac{5 / 13}{6 / 13}=\frac{5}{6} .
$$