9. Given the point (6,-8) values of the six trig function.10. Given that cot theta = -r3/2 in Quad II, find the state the six trig ratios. ​

Answer:Here's what I get.Step-by-step explanation:9. (6, -8)The reference angle θ is in the fourth quadrant.∆AOB is a right triangle.OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100OB = √100 = 10  [tex]\sin \theta = \dfrac{-8}{10} = -\dfrac{4}{5}\\\\\cos \theta =\dfrac{6}{10} = \dfrac{3}{5}\\\\\tan \theta = \dfrac{-8}{6} = -\dfrac{4}{3}\\\\\csc \theta = \dfrac{10}{-8} = -\dfrac{5}{4}\\\\\sec \theta = \dfrac{10}{6} = \dfrac{5}{3}\\\\\cot \theta = \dfrac{6}{-8} = -\dfrac{3}{4}[/tex]10. cot θ = -(√3)/2The reference angle θ is in the second quadrant.∆AOB is a right triangle.OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7OB = √7[tex]\sin \theta = \dfrac{2}{\sqrt{7}} = \dfrac{2\sqrt{7}}{7}\\\\\cos \theta = \dfrac{-\sqrt{3}}{\sqrt{7}} = -\dfrac{\sqrt{21}}{7}\\\\\tan \theta = \dfrac{2}{-\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\\\\\csc \theta = \dfrac{\sqrt{7}}{2} \\\\\sec \theta = \dfrac{\sqrt{7}}{-\sqrt{3}} = -\dfrac{\sqrt{21}}{3}\\\\\cot \theta = -\dfrac{\sqrt{3}}{2}[/tex]