Question

Given that θ terminates in Quadrant III and tanθ 5/12= , find cscθ

Answer 1

`\bf tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad csc(\theta)=\cfrac{hypotenuse}{opposite} \\\\ -------------------------------\\\\ tan(\theta)=\cfrac{5}{12}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}\\\\ -------------------------------\\\\ \textit{let's use the pythagorean theorem to get the hypotenuse`

now, the square root gives us the +/- version, so.. which is it? well, the hypotenuse is just a radius unit, is never negative, just the radius unit, so just the absolute value of that or the version 13, so c = 13

now, that we know what the hypotenuse "c" is, well


`\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)\cfrac{13}{5}`

Answer 2

`csc(\theta)=-(13)/(5)`

Explanation:

We've been told that θ terminates in Quadrant III, which means that θ is the angle between the positive x-axis and a vector
`\vec{r}` with negative x and y coordinates.

Additionally, we know that
`tan(\theta)=(5)/(12)=(y-coordinate)/(x-coordinate)`

So, using this information we can conclude that:

x-coordinate=
`x=-12`

y-coordinate=
`y=-5`

(This make sense because
`tan(\theta)=(-5)/(-12)=(5)/(12)`)

But, we still have to calculate
`csc(\theta)`

`csc(\theta)=(r)/(y-coordinate)`

Where
`r` is the vector magnitude

`r=√(x^2+y^2) =√((-12)^2+(-5)^2)=13`

`csc(\theta)=(13)/((-5))=-(13)/(5)`

So we've found the answer:
`csc(\theta)=-(13)/(5)`