## anonymous 5 years ago tan-1 (-1/square3)

You have to make the identification,$y=\tan^{-1}\left( -\frac{1}{\sqrt{3}} \right) \rightarrow \tan{y}=-\frac{1}{\sqrt{3}}$If you know your exact trig. solutions (in this case, those obtained by cutting an equilateral triangle in half), tan(y)=1/sqrt(3) when y = 30 degrees. But we want the negative situation. Tan is negative in the second and fourth quadrants, so our angle will lie there. For the second quadrant, we get, 180-30 degrees = 150 degrees = 5pi/3 radians and 360-30 degrees = 330 degrees = 11pi/6 radians assuming you only want solutions for 0 <= y < 2pi