The quantity \(\tan(7.5^\circ)\)can be expressed in the form \(\sqrt{a} + \sqrt{b} - \sqrt{c} - \sqrt{d}\) where a,b,c and d are positive integers. Enter a+b, c+d in that order.
tan (15°) = [1 -cos 30 ] / sin 30 = [ 1 -sqrt (3)/2 ] / (1/2) ] = [2 -sqrt (3)] / 1
sin (15°) = [ 2 -sqrt 3 ] / sqrt [ 1^2 + (2 -sqrt 3)^2 ] = [2 -sqrt 3 ] / sqrt [ 8 - 4sqrt 3]
cos (15°) = 1 / sqrt [ 8 - 4sqrt 3]
tan (7.5°) = [1 - cos 15 ] / sin 15 = [ 1 - 1 / sqrt [8 - 4sqrt 3 ] ] / ( [ 2 -sqrt 3] / sqrt [ 8 -4sqrt 3] ) =
sqrt [ 8 -4sqrt 3 ] * [ 1 - 1/ sqrt [ 8 -4sqrt 3] / [2 -sqrt 3] =
(sqrt [ 8 - 4sqrt 3 ] [ sqrt [8 - 4sqrt 3 ] - 1 ] / sqrt [8 -4sqrt 3]) / [2 -sqrt 3] =
(sqrt [ 8 -4sqrt 3 ] - 1) / [2 -sqrt 3 ] =
[ sqrt [ 8 - 4sqrt 3] - 1 ] [ 2 + sqrt 3 ] = { sqrt [ 8 - 4sqrt 3 ] = sqrt [( sqrt 6 - sqrt 2)^2] = sqrt 6 - sqrt 2 }
[(sqrt 6 - sqrt 2 ) -1 ] [ 2 + sqrt 3] =
2sqrt 6 - 2sqrt 2 - 2 + sqrt 18 - sqrt 6 - sqrt 3 =
sqrt 6 - 2sqrt 2- 2 + 3sqrt 2 -sqrt 3 =
sqrt 6 + sqrt 2 - sqrt 3 - 2 =
sqrt (2) + sqrt (6) -sqrt (3) - 2 =
sqrt (2) + sqrt (6) - sqrt (3) - sqrt (4)
a + b = 8
c + d = 7