Find cotθ if θ terminates in quadrant iii and secθ = - 2.

so hmmm let's recall, the hypotenuse, or radius, is never negative, since it's just a unit of the radius, so from -2/1 the negative fellow has to be the denominator, namely, the adjacent side, so a = -1, c = 2

bear also in mind, the angle is in the III quadrant, and on that quadrant, the sign for the x-coordinate is negative, so is -1

so.. what's the opposite side? or b? let's use the pythagorean theorem then



so.. which is it? the +/-?   well, we're on the III quadrant, on there both the "x" and "y" coordinates are both negative, so "b" has to also be negative, thus -√(3)=b