Let's begin by listing out the given information:
[tex]P(4,-2)[/tex]
We will solve using the formula:
[tex]\begin{gathered} r=\sqrt[]{x^2+y^2} \\ (x,y)=(4,-2) \\ r=\sqrt[]{4^2+(-2)^2}=\sqrt[]{16+4}=\sqrt[]{20} \\ r=\sqrt[]{20} \end{gathered}[/tex]
We will proceed to calculate for the six trigonometric function:
[tex]\begin{gathered} sin\theta=\frac{-2}{\sqrt[]{20}}=-\frac{2\sqrt[]{20}}{20}=\frac{\sqrt[]{20}}{10}=\frac{2\sqrt[]{5}}{10}=\frac{\sqrt[]{5}}{5} \\ sin\theta=\frac{\sqrt[]{5}}{5} \\ \\ cos\theta=\frac{4}{\sqrt[]{20}}=\frac{4\sqrt[]{20}}{20}=\frac{\sqrt[]{20}}{5} \\ cos\theta=\frac{\sqrt[]{20}}{5} \\ \\ tan\theta=\frac{-2}{4}=-\frac{2}{4}=-\frac{1}{2} \\ tan\theta=-\frac{1}{2} \end{gathered}[/tex]
For the remaining, we have:
[tex]\begin{gathered} cot\theta=\frac{1}{tan\theta}=\frac{1}{\frac{-1}{2}}=-\frac{2}{1} \\ cot\theta=-2 \\ \\ sec\theta=\frac{1}{cos\theta}=\frac{1}{\frac{\sqrt[]{20}}{5}}=\frac{5}{\sqrt[]{20}} \\ sec\theta=\frac{5}{\sqrt[]{20}} \\ \\ co\sec \theta=\frac{1}{sin\theta}=\frac{1}{\frac{\sqrt[]{5}}{5}}=\frac{5}{\sqrt[]{5}} \\ co\sec \theta=\frac{5}{\sqrt[]{5}} \end{gathered}[/tex]
