Answer
[tex]\begin{gathered} \tan \theta=-\frac{8}{15} \\ \cos ^{}\theta=\frac{15}{17} \end{gathered}[/tex]Explanation
Given:
[tex]\begin{gathered} \theta\text{ is in quadrant VI,} \\ \csc \theta=-\frac{17}{8} \end{gathered}[/tex]Using trigonometric identity, To find the exact value of tan θ
[tex]1+\cot ^2\theta=\csc ^2\theta[/tex][tex]\begin{gathered} 1+\cot ^2\theta=(-\frac{17}{8})^2 \\ 1+\cot ^2\theta=\frac{289}{64} \\ \cot ^2\theta=\frac{289}{64}-1 \\ \cot ^2\theta=\frac{289-64}{64} \\ \cot ^2\theta=\frac{225}{64} \\ \cot ^{}\theta=\pm\sqrt[]{\frac{225}{64}} \\ \cot \theta=\pm\frac{15}{8} \\ \text{But cot }\theta=\frac{1}{\tan \theta} \\ \Rightarrow\text{tan }\theta=\pm\frac{8}{15} \\ \text{Since }\theta\text{ is in quadrant VI, then tan }\theta\text{ is negative} \\ \tan \theta=-\frac{8}{15} \end{gathered}[/tex]To find the exact value of cos θ
[tex]\begin{gathered} \text{Using trigonometric identuty} \\ \cos ^2\theta=\frac{1}{1+\tan ^2\theta} \\ \cos ^2\theta=\frac{1}{1+(\frac{64}{225})} \\ \cos ^2\theta=\frac{1}{\frac{289}{225}} \\ \cos ^2\theta=\frac{225}{289} \\ \cos ^{}\theta=\pm\sqrt[]{\frac{225}{289}} \\ \cos ^{}\theta=\pm\frac{15}{17} \\ \text{Since }\theta\text{ is in quadrant VI, cos }\theta\text{ is positve} \\ \therefore\cos ^{}\theta=\frac{15}{17} \end{gathered}[/tex]θ