The terminal side of theta intersects with unit circle at (-√2/2, -√2/2)
The only quadrant where we have (-, -), that is the y and x coordinate is negative is the third quadrant
In the third quadrant: tan is positive, cos is negative and sine is negative
from the point of the unit circle:
[tex]\begin{gathered} x\text{ = adjacent = }\frac{-\sqrt[]{2}}{2} \\ y\text{ = opposite = }\frac{-\sqrt[]{2}}{2} \\ \text{hypotenuse}^2=opposite^2+adjacent^2 \\ \text{hypotenuse}^2=\text{ }(\frac{-\sqrt[]{2}}{2})^2\text{ + }(\frac{-\sqrt[]{2}}{2})^2 \\ \text{hypotenuse}^2=\frac{2}{4}\text{ + }\frac{2}{4}\text{ = }\frac{4}{4}=\text{ 1} \\ \text{hypotenuse = }\sqrt[]{1} \\ \text{hypotenuse = 1} \end{gathered}[/tex]We can find csc θ, sin θ and cot θ now:
[tex]\begin{gathered} co\sec \text{ }\theta\text{ = }\frac{1}{\sin \text{ }\theta} \\ \sin \text{ }\theta\text{ = }\frac{opposite}{hypotenuse} \\ \sin \theta\text{= }\frac{\frac{-\sqrt[]{2}}{2}}{1}\text{ = }\frac{-\sqrt[]{2}}{2} \\ \\ co\sec \text{ }\theta\text{ = }\frac{1}{\frac{-\sqrt[]{2}}{2}}\text{ = 1 }\div\text{ }\frac{-\sqrt[]{2}}{2}\text{ = 1 }\times\text{ }\frac{2}{-\sqrt[]{2}} \\ co\sec \text{ }\theta\text{ = }\frac{2}{-\text{ }\sqrt[]{2}} \end{gathered}[/tex][tex]\begin{gathered} \sin \theta\text{ = }\frac{opposite}{\text{hypotenuse}} \\ \sin \theta\text{ = }\frac{-\text{ }\sqrt[]{2}}{2} \end{gathered}[/tex][tex]\begin{gathered} \cot \theta\text{ = }\frac{1}{\tan\theta\text{ }} \\ \tan \text{ }\theta\text{ = }\frac{\sin\theta}{\cos\theta} \\ \\ \cos \text{ }\theta\text{ = }\frac{adjacent}{\text{hypotenuse}} \\ \cos \text{ }\theta\text{ = }\frac{\frac{-\sqrt[]{2}}{2}}{1}\text{ = }\frac{-\sqrt[]{2}}{2} \\ sin\text{ }\theta\text{ = }\frac{-\sqrt[]{2}}{2} \\ \\ \tan \text{ }\theta\text{ = }\frac{\frac{-\sqrt[]{2}}{2}}{\frac{-\sqrt[]{2}}{2}}\text{ = 1} \\ \cot \theta=\frac{1}{1} \\ \cot \theta\text{ = 1} \end{gathered}[/tex]