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.**

tan(x+pi/3) = (tanx + tan pi/3)/(1-tanx.tan pi/6)

we know that tan pi/6 = √3 , tanx=1/2

= (1/2+√3) / (1-√3/2)

= [(1+2√3)/2] / [(2-√3)/2]

= (1+2√3)/(2-√3)

multiply the numerator and denominator by (2+√3)

= (1+2√3)(2+√3)/(2-√3)(2+√3)

= (2+5√3 +6)/ (4-3)

= 8 +5√3