Given:
[tex]\cot \theta=-\frac{2}{17}[/tex]
And θ is an angle in the standard position whose terminal side is in Quadrant IV.
We know that,
[tex]\cot \theta=\frac{\text{adjacent side}}{\text{opposite side}}[/tex]
Hence, opposite side=17 and adjacent side=2
Using pythogoras theorem,
[tex]\begin{gathered} \text{Hyp}^2=\text{Opp}^2+\text{Adj}^2 \\ \text{Hyp}^2=17^2+2^2 \\ \text{Hyp}^2=289+4 \\ \text{Hyp}^2=293 \\ \text{Hyp}=\sqrt[]{293} \end{gathered}[/tex]
Then the other values are,
Since θ lies in IV quadrant.
cos and sec values are positve and the others have negative values.
So,
[tex]\begin{gathered} \sin \theta=-\frac{opp}{hyp} \\ =-\frac{17}{\sqrt[]{293}} \\ \cos \theta=\frac{adj}{hyp} \\ =\frac{2}{\sqrt[]{293}} \\ t\text{an }\theta=-\frac{\text{opp}}{\text{adj}} \\ =-\frac{17}{2} \\ co\sec \theta=-\frac{hyp}{opp} \\ =-\frac{\sqrt[]{293}}{17} \\ \sec \theta=\frac{hyp}{\text{adj}} \\ =\frac{\sqrt[]{293}}{2} \end{gathered}[/tex]
Hence, the correct option is B.
