Question

If sec= 5/3 and the terminal point determined by is in quadrant 4, then

Answer 1

`\bf sec(\theta)=\cfrac{1}{cos(\theta)}\\\\ -----------------------------\\\\ sec(\theta)=\cfrac{5}{3}\implies \cfrac{1}{cos(\theta)}=\cfrac{5}{3}\implies \cfrac{3}{5}=cos(\theta) \\\\\\ \textit{now, }\theta\textit{ is in the 4th quadrant}`

so... notice the picture below, the angle in the fourth quadrant
now... notice, the cosine is just the distance the angle makes with the x-axis
so... using the pythagorean theorem, get the opposite side, from that triangle, keeping in mind that, "y" is negative in the 4th quadrant

once you get "y", get any other trigonometry value you need :)

Answer 2

C.
`csc(\theta)=-(5)/(4)`

D.
`cos(\theta)= (3)/(5)`

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:

`I=First\hspace{3} quadrant\\II=Second\hspace{3} quadrant\\III=Third\hspace{3} quadrant\\IV=Fourth\hspace{3} quadrant`

`a=Opposite\\b=Adjacent\\c=Hypotenuse`

Now, the functions on a right triangle like this are given by:

`sin(\theta)=(a)/(c)\hspace{15}csc(\theta)=(c)/(a) \\cos(\theta)=(b)/(c)\hspace{15}sec(\theta)=(c)/(b) \\tan(\theta)=(a)/(b)\hspace{15}cot(\theta)=(b)/(a)`

If:

`sec(\theta)=(5)/(3)`

Then:

`c=5\\b=3`

Using pythagorean theorem we can find a:

`c^2=a^2+b^2`

Solving for a:

`a^2=5^2-3^2=16\\a=√(16) =\pm4`

Since the triangle is in the fourth quadrant, the value of the opposite angle is negative, hence:

`a=-4`

Now, let's check every option, so we can determinate which of them are true or false.

A.
`tan(\theta)=(4)/(3)`

`tan(\theta)=(a)/(b) =(-4)/(3) \\eq (4)/(3)`

This is incorrect.

B.
`sin(\theta)=-(2)/(5)`

`sin(\theta)=(a)/(c) =(-4)/(5) \\eq - (2)/(5)`

This is incorrect.

C.
`csc(\theta)=-(5)/(4)`

`csc(\theta)=(c)/(a) =(5)/(-4) =- (5)/(4)`

This is correct

D.
`cos(\theta)= (3)/(5)`

`cos(\theta)=(b)/(c) =(3)/(5)`

This is correct.