Answer: \(\cos\theta=-\sqrt{\frac56}\), \(\sin\theta=-\sqrt{\frac16}\), and \(\sec\theta+\cot\theta=\frac{5-\sqrt6}{\sqrt5}\).

Use the standard trigonometric identities and make sure the final angles are chosen from the interval given in the question.

Since \(\tan\theta\) is positive and \(180^\circ\lt \theta\lt 360^\circ\), \(\theta\) is in the third quadrant. Therefore both \(\sin\theta\) and \(\cos\theta\) are negative.

Now

\(\tan\theta=\frac{\sqrt5}{5}=\frac{1}{\sqrt5}.\)

Using a reference triangle, take opposite side \(1\), adjacent side \(\sqrt5\), and hypotenuse \(\sqrt6\).

Thus

\(\cos\theta=-\frac{\sqrt5}{\sqrt6}=-\sqrt{\frac56},\)

and

\(\sin\theta=-\frac{1}{\sqrt6}=-\sqrt{\frac16}.\)

Also

\(\sec\theta+\cot\theta=\frac{1}{\cos\theta}+\frac{1}{\tan\theta}.\)

So

\(\sec\theta+\cot\theta=-\frac{\sqrt6}{\sqrt5}+\sqrt5=\frac{5-\sqrt6}{\sqrt5}.\)

This gives the required answer: \(\cos\theta=-\sqrt{\frac56}\), \(\sin\theta=-\sqrt{\frac16}\), and \(\sec\theta+\cot\theta=\frac{5-\sqrt6}{\sqrt5}\).