Answer :
We know that the angle terminates in the fourth quadrant and that the tangent of it is -3/4.
Angles that are on the fourth quadrant have a negative sine and a positive cosine.
We can start with the cot():
[tex]\cot\theta=\frac{1}{\tan\theta}=\frac{1}{-\frac{3}{4}}=-\frac{4}{3}[/tex]We can relate the cosine with the tangent as:
[tex]\begin{gathered} \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{1-\cos^2\theta}}{\cos\theta} \\ -\frac{3}{4}=\frac{\sqrt{1-\cos^2\theta}}{\cos\theta} \\ -\frac{3}{4}\cos\theta=\sqrt{1-\cos^2\theta} \\ (-\frac{3}{4}\cos\theta)^2=1-\cos^2\theta \\ \frac{9}{16}\cos^2\theta=1-\cos^2\theta \\ (\frac{9}{16}+1)\cos^2\theta=1 \\ \frac{9+16}{16}\cos^2\theta=1 \\ \frac{25}{16}\cos^2\theta=1 \\ \cos^2\theta=\frac{16}{25} \\ \cos\theta=\sqrt{\frac{16}{25}} \\ \cos\theta=\frac{4}{5} \end{gathered}[/tex]We can now calculate the sine of the angle as:
[tex]\begin{gathered} \sin\theta=\tan\theta\cdot\cos\theta \\ \sin\theta=-\frac{3}{4}\cdot\frac{4}{5}=-\frac{3}{5} \end{gathered}[/tex]We can now calculate the sec() and csc() as:
[tex]\begin{gathered} \sec\theta=\frac{1}{\cos\theta}=\frac{5}{4} \\ \\ \csc\theta=\frac{1}{\sin\theta}=-\frac{5}{3} \end{gathered}[/tex]Answer:
sin() = -3/5
cos() = 4/5
cot() = -4/3
sec() = 5/4
csc() = -5/3
