Answer:
C. [tex]csc(\theta)=-\frac{5}{4}[/tex]
D. [tex]cos(\theta)= \frac{3}{5}[/tex]
Step-by-step explanation:
First, take a look to the picture I attached you. I drew a diagram according to the data provided from the problem. From this diagram:
[tex]I=First\hspace{3} quadrant\\II=Second\hspace{3} quadrant\\III=Third\hspace{3} quadrant\\IV=Fourth\hspace{3} quadrant[/tex]
[tex]a=Opposite\\b=Adjacent\\c=Hypotenuse[/tex]
Now, the functions on a right triangle like this are given by:
[tex]sin(\theta)=\frac{a}{c}\hspace{15}csc(\theta)=\frac{c}{a} \\cos(\theta)=\frac{b}{c}\hspace{15}sec(\theta)=\frac{c}{b} \\tan(\theta)=\frac{a}{b}\hspace{15}cot(\theta)=\frac{b}{a}[/tex]
If:
[tex]sec(\theta)=\frac{5}{3}[/tex]
Then:
[tex]c=5\\b=3[/tex]
Using pythagorean theorem we can find a:
[tex]c^2=a^2+b^2[/tex]
Solving for a:
[tex]a^2=5^2-3^2=16\\a=\sqrt{16} =\pm4[/tex]
Since the triangle is in the fourth quadrant, the value of the opposite angle is negative, hence:
[tex]a=-4[/tex]
Now, let's check every option, so we can determinate which of them are true or false.
A. [tex]tan(\theta)=\frac{4}{3}[/tex]
[tex]tan(\theta)=\frac{a}{b} =\frac{-4}{3} \neq \frac{4}{3}[/tex]
This is incorrect.
B. [tex]sin(\theta)=-\frac{2}{5}[/tex]
[tex]sin(\theta)=\frac{a}{c} =\frac{-4}{5} \neq - \frac{2}{5}[/tex]
This is incorrect.
C. [tex]csc(\theta)=-\frac{5}{4}[/tex]
[tex]csc(\theta)=\frac{c}{a} =\frac{5}{-4} =- \frac{5}{4}[/tex]
This is correct
D. [tex]cos(\theta)= \frac{3}{5}[/tex]
[tex]cos(\theta)=\frac{b}{c} =\frac{3}{5}[/tex]
This is correct.
