How do you evaluate #cot ((7pi)/6)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Konstantinos Michailidis Jul 9, 2016 It is #cot((pi+6pi)/6)=cot(pi+pi/6)=cot(pi/6)=sqrt3# Another way is that #cot(pi+pi/6)=1/(tan(pi+pi/6))# Use the #tan(a + b) = [tan a + tan b]/[1 - (tan a)(tan b)] # so you get: #tan(pi + pi/6) = [tanpi + tanpi/6]/[1 - (tanpi)(tanpi/6)] = [ 0 + (sqrt3)/3] / [1 - (0)((sqrt3)/3)] = [(sqrt3)/3] / [1 - 0] = [(sqrt3)/3] / 1 = [sqrt3] / 3 # Hence #cot(pi+pi/6)=1/(sqrt3/3)=sqrt3# 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 20703 views around the world You can reuse this answer Creative Commons License