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tan^-1 -root3
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\(\large tan^{-1}(-\sqrt3)=\theta \) is equivalent to \(\large tan\theta = -\sqrt3 \). if you can answer the second question, you'll have answered the first.
This is asking you "What angle will give you a tangent of -root3? Since the tangent is sin/cos or y/x, tangent is negative in the second and fourth quadrants. Since you are told that y = -root3 and it is understood that x = 1, our angle is in the fourth quadrant where y is negative and x is positive. The side across from the angle is -root3, so which of our special angles has a tangent of root 3/1?
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