How do you evaluate #tan(-pi/6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Konstantinos Michailidis Mar 23, 2016 We have that #tan(-pi/6)=sin(-pi/6)/cos(-pi/6)=-sin(pi/6)/cos(pi/6)=-[1/2]/[sqrt3/2]=-sqrt3/3# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 9600 views around the world You can reuse this answer Creative Commons License