Question 1139929
<font face="times" color="black" size="3">
Part A


Answer: <font color=red size=4>True</font>


-------------------------------------------------------
Explanation:


One way to define the tangent ratio is 
*[Tex \Large \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}]


We can also say
*[Tex \Large \tan(\theta) = \frac{y}{x}]
where y = sin(theta) and x = cos(theta)


In quadrant II, sine is positive while cosine is negative. The ratio of the two will ultimately be negative since the rule "positive divided by negative = negative" comes into play. For example, 10/(-2) = -5. 


So a statement like *[Tex \Large \tan(\theta) = -1.323] is possible when theta is in quadrant 2. 


Here's another way to think of it: the tangent ratio is the slope of the line connecting the origin point to the terminal point on the unit circle. In quadrant II, this slope is negative as it goes downhill when we read from left to right. See the diagram at the bottom of the page. 


====================================================================================================
Part B


Answer: <font color=red size=4>False</font>


-------------------------------
Explanation:


We'll use the fact that sine is positive in quadrant II to directly jump to cosecant (csc) is also positive. 


The reciprocal trig identity
*[Tex \Large \csc(\theta) = \frac{1}{\sin(\theta)}]
helps us see the connection. 


Therefore, saying something like *[Tex \Large \csc(\theta) = -0.542] is NOT possible due to theta being in quadrant II. It would only be possible if theta was in quadrant III or quadrant IV, which is where sine is negative. Sine is negative here because the y coordinate of the point on the unit circle is negative.


====================================================================================================


Diagram:
<img src = "https://i.imgur.com/NVEb1Zq.png">
The blue circle is the unit circle, which has a radius of 1 and it is centered at the origin. Point A is some point on the unit circle such that it is in quadrant II. This point is allowed to move around as long as it stays in quadrant II only. Point B is at the origin. The segment AB is shown in red. This segment has a negative slope because we go downhill as we move our eyes from left to right. So this helps confirm why the tangent ratio is negative. Furthermore, point A has a positive y coordinate, so the sine ratio and cosecant ratio will be positive, helping confirm that cosecant is never negative here. 
</font>