How do you simplify #tan (-5pi / 6) - cot (7pi / 2) #? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nghi N. Nov 20, 2015 Simplify trig expression. Ans: #sqrt3/3# Explanation: #tan #(-5pi)/6# = tan (pi/6 - (6pi)/6) = tan (pi/6 - pi) = tan (pi/6) = # # = sqrt3/3.# #cot ((7pi)/2) = cot ((3pi)/2 + 2pi) = cot ((3pi)/2) = 0#, then #tan ((-5pi)/6) - cot ((7pi)/2) = sqrt3 - 0 = 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 3328 views around the world You can reuse this answer Creative Commons License