How do you evaluate $cot[arccos(-1/2)+arccos(0)+arctan(1/sqrt(3))]$ ?

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Answer 1 · 2016-07-04T15:00:00

$1/sqrt 3$ .

A trigonometric function is the same for any general value. So, only

principal values are considered.

$a = arc cos (-1/2). cos a = (-1/2)<0.$

a = angle in the 2nd quadrant.. .

So, a = $(2pi)/3$

$b = arc cos 0 = pi/2$ .

$c = arc tan (1/sqrt 3 )= pi/6$ .

So, the given expression

$= cot ( a + b + c ) =cot((2pi)/3+ pi/2+pi/6)$

$= cot ((4pi)/3) = cot (pi+pi/3)=cot (pi/3)$

$=1/sqrt 3$ .

. .

How do you evaluate cot[arccos(-1/2)+arccos(0)+arctan(1/sqrt(3))]cot[arccos(-1/2)+arccos(0)+arctan(1/sqrt(3))]?