Find the exact value of cot Ø if csc Ø = -4/3 and and the terminal side of Ø lies in Quadrant III

Answer:
[tex]cot \theta= \frac{\sqrt{7} }{3}[/tex]
Step-by-step explanation:
It is given that cosec [tex]\theta[/tex]=-4/3 and sin theta =1/cosec theta
so sin theta =-3/4 now since sin theta is perpendicular/hypotenuse, the third side base will be [tex]\sqrt({4}^{2}- (-3)^{2})\\ =\sqrt{16-9}\\ =\sqrt{7}[/tex]
and we know that cot theta is inverse of tan so it will be base/perpendicular
i.e [tex]\frac{\sqrt{7} }{3}[/tex] and also it is given that side of theta is in quadrant III so cot theta will be positive, so final answer will be
[tex]\frac{\sqrt{7} }{3}[/tex]
Answer:
√7/3.
Step-by-step explanation:
csc Ø = 1 / sin Ø = 1 / -4/3 = -3/4.
As the sine is the opposite / hypotenuse the side adjacent to angle Ø will have length √(4^2 - 3^2) = √7 (by the Pythagoras theorem) and as the angle is in the third quadrant this will be -√7.
So cot Ø = adjacent / opposite = -√7/ -3
= √7/3 (answer).