How do you evaluate #cot (pi/2 + 11pi/6) #?

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Answer 1 · 2016-07-04T07:46:32

#cot(pi/2+(11pi)/6)=1/sqrt3#

It is observed that #(11pi)/6# is just less than #2pi# and all trigonometric ratios get repeated after #2pi#, Hence, we can have

#cot(pi/2+(11pi)/6)#

= #cot(pi/2+(12pi)/6-pi/6)#

= #cot(pi/2+2pi-pi/6)#

= #cot(pi/2-pi/6)#

Further as cotangent of a complementary angle and its tangent are equal, we have

#cot(pi/2-pi/6)#

= #tanpi/6#

= #1/sqrt3#