SOLUTION
Let us make a simple diagram to illustrate the question and see the position of the angle
From the right-angled triangle in the diagram, the opposite side is 7, and the adjacent side is 9, we need to find the hypotenuse h.
From Pythagoras theorem,
[tex]\begin{gathered} \text{hyp}^2=opp^2+adj^2 \\ h^2=7^2+9^2 \\ h^2=49+81 \\ h=\sqrt[]{130} \end{gathered}[/tex]
Now we know that
[tex]\begin{gathered} \sin \theta=\frac{opposite}{\text{hypotenuse}} \\ \sin \theta=\frac{7}{\sqrt[]{130}} \\ This\text{ position of the angle }\theta\text{ is the 4th quadrant.} \\ In\text{ the 4th quadrant, only cos }\theta\text{ is positive, sin }\theta\text{ and tan }\theta\text{ are negative.} \\ \text{Hence } \\ \sin \theta=-\frac{7}{\sqrt[]{130}} \end{gathered}[/tex]
Also
[tex]\begin{gathered} co\sec \theta=\frac{1}{\sin \theta} \\ co\sec \theta=-\frac{\sqrt[]{130}}{7} \end{gathered}[/tex]
Hence, the answer is
[tex]-\frac{\sqrt[]{130}}{7}[/tex]
