We have an angle theta and we have to calculate the six trigonometric functions.
We know that:
[tex]\begin{gathered} \sin \theta=\frac{2}{3} \\ \tan \theta<0 \end{gathered}[/tex]
As the sine is positive and the tangent is negative, we can conclude that the cosine is negative.
We can find the cosine of theta using the following identity:
[tex]\begin{gathered} \sin ^2\theta+\cos ^2\theta=1 \\ \cos ^2\theta=1-\sin ^2\theta \\ \cos ^2\theta=1-(\frac{2}{3})^2 \\ \cos ^2\theta=1-\frac{4}{9} \\ \cos ^2\theta=\frac{9-4}{9} \\ \cos ^2\theta=\frac{5}{9} \\ \cos \theta=-\frac{\sqrt[]{5}}{3} \end{gathered}[/tex]
Now that we know the sine and the cosine, we can derive all the other trigonometric functiosn as:
[tex]\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{2}{3}}{-\frac{\sqrt[]{5}}{3}}=-\frac{2}{\sqrt[]{5}}=-\frac{2\sqrt[]{5}}{5}[/tex][tex]\cot \theta=\frac{1}{\tan\theta}=-\frac{\sqrt[]{5}}{2}[/tex][tex]\csc \theta=\frac{1}{\sin \theta}=\frac{3}{2}[/tex][tex]\sec \theta=\frac{1}{\cos\theta}=-\frac{3}{\sqrt[]{5}}=-\frac{3\sqrt[]{5}}{5}[/tex]
Answer:
cos θ = -√5/3
tan θ = -2√5/5
cot θ = -√5/2
csc θ = 3/2
sec θ = -3√5/5
