Find the equation of the tangent line to the graph of the function

f(x)=tan x at x=pi/3
So I'm using the y1-y2=m(x1-x2) formula.
Then I find the derivative of f(x)...f'(x)=sec^2(x)
Then I plug in pi/3 to x to get 4.
That should then be my slope which is 4.
I am stuck here, what else do I do? "/

y=tanx
dy=(secx)^2
(sec(pi/3))^2 = 4 = slope
point is (pi/3,tan(pi/3)) or (pi/3,sqrt3)
So point-slope form is (y-sqrt3) = 4(x-pi/3)
Solve for y: y=4x - 4pi/3 +sqrt3