Solution:-
To evaluate [tex]tan^{-1}(-\sqrt{3} )[/tex]
Consider a right angle triangle with base angle x =[tex]tan^{-1}(-\sqrt{3} )[/tex]
⇒tan x = [tex](-\sqrt{3} )[/tex]
We know that for[tex]\theta[/tex]=60°, tan(60°)=[tex]\sqrt{3}[/tex]
here the value is [tex](-\sqrt{3} )[/tex] so it will lie in 2nd or 4th quadrant.
For 2nd quadrant the value of x will be x=180°-[tex]\theta[/tex]
⇒ x=180°-60°=120°.
For 4th quadrant the value of x will be x=360°-[tex]\theta[/tex]
⇒ x=360°-60°=300°.
In radian
For 2nd quadrant the value of x will be x=180°-[tex]\theta[/tex]
⇒[tex]x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}[/tex] =1.04(approx)
For 4th quadrant the value of x will be x=360°-[tex]\theta[/tex]
⇒[tex]x=2\pi-\frac{\pi}{3}=\frac{5\pi}{3}[/tex] .
