Question 1198723
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In response to what the tutor @MathLover1 wrote: Cotangent is NOT negative in quadrant Q3.
Take note how sine and cosine are indeed negative here, so,
cot = cos/sin
the two negative items divide to a positive result.


In short, cotangent is positive in Q3.


Since cot(theta) = -2 is negative, this places theta in either Q2 or Q4.


Then we're told csc(theta) < 0, which fully narrows things down to Q4 only.
270 < theta < 360 in degree mode
3pi/2 < theta < 2pi in radian mode
The actual theta value itself doesn't matter; however, this interval helps narrow things down a bit.


Since cot(theta) = -2 = 2/(-1), we could have a triangle with adjacent side 2 and opposite side -1.
I'll make the opposite length "negative" so that we can keep the proper signs in mind. Of course a negative length isn't possible. 
It's purely as a means to retain information.
I'm making the 1 negative so that it indicates we're below the x axis, i.e. the y coordinate is negative.


Use the pythagorean theorem to find the hypotenuse is sqrt(5) units long.


This is one way to draw the triangle in Q4
*[illustration Screenshot_169.png]
We have:
opposite = -1
adjacent = 2
hypotenuse = sqrt(5)


Then recall that
sin(theta) = opposite/hypotenuse
sec(theta) = hypotenuse/adjacent


I'll let you finish up from here.
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