Answer:
See below
Step-by-step explanation:
We have that: [tex]\cot \theta=-\frac{2}{17}[/tex], then the reciprocal is [tex]\tan \theta=-\frac{17}{2}[/tex]
Applying the Pythagoras, we have [tex]h^2=17^2+2^2[/tex]
[tex]h^2=289+4=293[/tex]
[tex]h=\sqrt{293}[/tex]
Sine is opposite/hypotenuse
[tex]\sin \theta=\frac{2}{\sqrt{293}}=\frac{2\sqrt{293}}{293}[/tex]
Cosine is adjacent/hypotenuse
[tex]\cos \theta=\frac{17}{\sqrt{293}}=\frac{17\sqrt{293}}{293}[/tex]
The reciprocal of sine
[tex]\csc \theta=\frac{\sqrt{293}}{2}[/tex]
The reciprocal of cosine is
[tex]\sec \theta=\frac{\sqrt{293}}{17}[/tex]
