Respuesta :
Answer::
[tex]\begin{gathered} \sin (\theta)=\frac{\sqrt[]{5}}{3},\textcolor{red}{tan(\theta)=-\frac{\sqrt[]{5}}{2},}\cot (\theta)=-\frac{2\sqrt[]{5}}{5} \\ \sec (\theta)=-\frac{3}{2},\textcolor{red}{\csc (\theta)=\frac{3\sqrt[]{5}}{5}} \end{gathered}[/tex]Explanation:
Given:
[tex]\begin{gathered} \cos \theta=-\frac{2}{3} \\ \tan \theta<0 \end{gathered}[/tex]If the cosine and tangent of an angle are both negative, then the angle is in Quadrant II.
First, determine the length of the opposite side using the Pythagorean Theorem.
[tex]\begin{gathered} \cos \theta=\frac{-2}{3}\implies\text{Adjacent}=-2,\; \text{Hypotenuse}=3 \\ \text{Hyp}^2=\text{Adj}^2+\text{Opp}^2 \\ 3^2=(-2)^2+\text{Opp}^2 \\ \text{Opp}^2=9-4=5 \\ \text{Opposite}=\sqrt{5} \end{gathered}[/tex]The length of the opposite side is √5.
Therefore:
[tex]\begin{gathered} \sin (\theta)=\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac{\sqrt[]{5}}{3} \\ \tan (\theta)=\frac{\text{Opposite}}{\text{Adjacent}}=-\frac{\sqrt[]{5}}{2} \end{gathered}[/tex]Cotangent is the inverse of tangent, therefore:
[tex]\begin{gathered} \cot (\theta)=\frac{1}{\tan(\theta)}=-\frac{2}{\sqrt[]{5}} \\ \text{Rationalise the denominator} \\ =-\frac{2}{\sqrt[]{5}}\times\frac{\sqrt[]{5}}{\sqrt[]{5}} \\ \implies\cot (\theta)=-\frac{2\sqrt[]{5}}{5} \end{gathered}[/tex]Secant is the inverse of Cosine, therefore:
[tex]\sec (\theta)=\frac{1}{\cos(\theta)}=-\frac{3}{2}[/tex]Cosecant is the inverse of Sine, therefore:
[tex]\begin{gathered} \csc (\theta)=\frac{1}{\sin(\theta)}=\frac{3}{\sqrt[]{5}} \\ \text{Rationalise the denominator} \\ =\frac{3}{\sqrt[]{5}}\times\frac{\sqrt[]{5}}{\sqrt[]{5}} \\ \implies\csc (\theta)=\frac{3\sqrt[]{5}}{5} \end{gathered}[/tex]