How do you use the angle sum or difference identity to find the exact value of #tan(pi/4+pi/3)#?

From a #1:1:sqrt(2)# right angled triangle (half of a square), we find that #tan(pi/4) = 1#

From a #1:sqrt(3):2# right angled triangle (half of an equilateral triangle), we find that #tan(pi/3) = sqrt(3)#

The sum formulae for #sin# and #cos# are:

#sin(alpha+beta) = sin(alpha)cos(beta)+sin(beta)cos(alpha)#

#cos(alpha+beta) = cos(alpha) cos(beta) - sin(alpha)sin(beta)#

From these we can deduce the sum formula for #tan# (in case you did not know it already):

#tan(alpha+beta) = sin(alpha+beta)/cos(alpha+beta)#

#color(white)(tan(alpha+beta)) = (sin alpha cos beta + sin beta cos alpha)/(cos alpha cos beta - sin alpha sin beta)#

#color(white)(tan(alpha+beta)) = ((sin alpha cos beta + sin beta cos alpha) -: (cos alpha cos beta))/((cos alpha cos beta - sin alpha sin beta) -: (cos alpha cos beta))#

#color(white)(tan(alpha+beta)) = (tan alpha + tan beta)/(1 - tan alpha tan beta)#

Hence we have:

#tan (pi/4 + pi/3) = (tan (pi/4) + tan (pi/3)) / (1 - tan (pi/4) tan (pi/3))#

#color(white)(tan (pi/4 + pi/3)) = (1 + sqrt(3)) / (1 - sqrt(3))#

#color(white)(tan (pi/4 + pi/3)) = ((1 + sqrt(3))(1+sqrt(3))) / ((1 - sqrt(3))(1+sqrt(3)))#

#color(white)(tan (pi/4 + pi/3)) = (1 + 2sqrt(3) + 3) / (1 - 3)#

#color(white)(tan (pi/4 + pi/3)) = (4 + 2sqrt(3)) / (-2)#

#color(white)(tan (pi/4 + pi/3)) = -2-sqrt(3)#