Respuesta :
Given that:
5π/3
=5/3×180
=240°
This is in the Quadrant III
The value of cot 5π/3 will be given by:
cost θ=1/tan θ
thus
tan θ
=tan 5π/3
=tan 240
=tan (240-180)
=tan 60
using an equilateral triangle with lengths 2 units
then
tan 60=√3/1
tan 60=√3
hence
tan 240=√3
5π/3
=5/3×180
=240°
This is in the Quadrant III
The value of cot 5π/3 will be given by:
cost θ=1/tan θ
thus
tan θ
=tan 5π/3
=tan 240
=tan (240-180)
=tan 60
using an equilateral triangle with lengths 2 units
then
tan 60=√3/1
tan 60=√3
hence
tan 240=√3
Answer:
In Quadrant I(0°≤A≤90°)= All trigonometric functions are positive
In Quadrant II(90°≤A≤180°)=Sine and Cosine are positive.
In Quadrant III(180°≤A≤270°)=Cot and Tan are positive.
In Quadrant IV(270°≤A≤360°)=Sec and Cos are positive.
[tex]Cot (\frac{5\pi}{3})=Cot (2\pi -\frac{\pi}{3})\\\\Cot (2\pi -A^{\circ})=-Cot A^{\circ}\\\\Cot (2\pi -\frac{\pi}{3})= -Cot (\frac{\pi}{3})\\\\Cot(\frac{180^{\circ}}{3})=-Cot 60^{\circ}\\\\-Cot 60^{\circ}=\frac{-1}{\sqrt{3}}[/tex]