A tangent or slope of #1/2# is a right triangle with opposite #1#, adjacent #2# so a hypotenuse of #sqrt{1^2+2^2}=sqrt{5}.# The other trig functions are made of these components. We can't be sure of the sign of any of them except the cotangent.
#cos arctan(1/2) = pm 2/sqrt{5}#
#sin arctan(1/2) = pm 1/sqrt{5}#
# tan arctan(1/2) = 1/2 #
#cot arctan(1/2) = 1/{tan arctan(1/2)} = 2 #
#csc arctan(1/2) = pm sqrt{5} #
#sec arctan(1/2) = pm \sqrt{5}/2#
Let's work out six trig functions in general given #tan x = a/b.# That's a right triangle, opposite #a#, adjacent #b#, hypotenuse #sqrt{a^2+b^2}.#
#cos arctan(a/b) = pm b/sqrt{a^2+b^2}#
#sin arctan(a/b) = pm a/sqrt{a^2+b^2} #
#tan arctan(a/b) = a/b #
#cot arctan(a/b) = b/a #
#csc arctan(a/b) = pm sqrt{a^2+b^2}/a#
#sec arctan(a/b) = pm sqrt{a^2+b^2}/b#
If they have a square root they get a #pm#.