sec⁡θ=−3,tan⁡θ>0. Find the exact value of the remaining trigonometric functions of θ.

We have given the value of a trigonometric function ${\displaystyle \sec \left(\theta \right)}$ and asked the values of other remaining trigonometric functions. So here we will use some standard identities and results in order to determine the values of other trigonometric functions.
${\displaystyle {\sec }^2\theta -{\tan }^2\theta =1}$
${\displaystyle \cos \theta =\frac{1}{\sec \theta }}$
${\displaystyle \sin \theta =\frac{\tan \theta }{\sec \theta }}$
${\displaystyle \cos ec\theta =\frac{1}{\sin \theta }}$
${\displaystyle \cot \theta =\frac{1}{\tan \theta }}$
Since, ${\displaystyle \sec \theta =-3<0\text{and}\tan \theta >0}$ then ${\displaystyle \theta }$ must be in third quadrant in which ${\displaystyle \sin \theta ,\cos \theta ,\cos ec\theta \text{and}\sec \theta }$ will be negative and rest ${\displaystyle \tan \theta \text{and}\cot \theta }$ will be positive.
Now,
${\displaystyle {\sec }^2\theta -{\tan }^2\theta =1}$
${\displaystyle {(-3)}^2-{\tan }^2\theta =1}$
${\displaystyle {\tan }^2\theta ={(-3)}^2-1}$
${\displaystyle \tan \theta =\sqrt{8}}$
Also,
${\displaystyle \cos \theta =\frac{1}{\sec \theta }=\frac{1}{-3}=-\frac{1}{3}}$
${\displaystyle \sin \theta =\frac{\tan \theta }{\sec \theta }=\frac{\sqrt{8}}{-3}=\frac{-2\sqrt{2}}{3}}$
${\displaystyle \cos ec\theta =\frac{1}{\sin \theta }=\frac{1}{\frac{-2\sqrt{2}}{3}}=-\frac{3}{2\sqrt{2}}}$
${\displaystyle \cot \theta =\frac{1}{\tan \theta }=\frac{1}{\sqrt{8}}=\frac{1}{2\sqrt{2}}}$
Hence, values of other trigonometric functions ${\displaystyle \sin \theta ,\cos \theta ,\tan \theta \cos ec\theta \text{and}\cot \theta }$ will be ${\displaystyle \frac{-2\sqrt{2}}{3},-\frac{1}{3},2\sqrt{2},-\frac{3}{2\sqrt{2}}\text{and}\frac{1}{2\sqrt{2}}}$ respectively.

Answer is given below (on video)

Answer is given below (on video)