If tanx=1/2 what is tan(x+pi/3)

We'll recall the following identity concerning the tangent
of the sum of two angles:

tan(a+b) = (tan a + tan
b)/[1-(tan a)*(tan b)]

Comparing, we'll
get:

tan (x + pi/3) = (tan x + tan pi/3)/[1-(tan x)*(tan
pi/3)]

But tan pi/3 = sqrt3 and tan x = 1/2, therefore,
we'll have;

tan (x + pi/3) = (1/2 +
sqrt3)/[1-(1/2)*(sqrt3)]

tan (x + pi/3) = [(1 +
2sqrt3)/2]/[(2-sqrt3)/2]

tan (x + pi/3) = (1 +
2sqrt3)/(2-sqrt3)

We'll multiply by the conjugate of
denominator:

tan (x + pi/3) = (1 +
2sqrt3)(2+sqrt3)/(2-sqrt3)(2+sqrt3)

The product from
denominator returns the difference of two
squares:

(2-sqrt3)(2+sqrt3) = 4 -
3

tan (x + pi/3) = (1 +
2sqrt3)(2+sqrt3)/(4-3)

tan (x + pi/3) = 2 + sqrt3 + 4sqrt3
+ 6

tan (x + pi/3) = 8 +
5sqrt3

The tangent of the sum x + pi/3 is tan
(x + pi/3) = 8 + 5sqrt3.