Maths problem

(1)
cos^2 (90 - a) - sin^2(90 - a)
= sin^a - cos^a
= (1 - cos^2a) - cos^2a
= 1 - 2 cos^2 a.
(2)
sin ( x + 50) + sin(40 - x)tan(x + 50) = sqrt 3
sin[90 - (40 - x)] + sin ( 40 - x) tan [ 90 - (40 - x)] = sqrt 3
cos(40 - x) + sin(40 - x)/tan ( 40 - x) = sqrt 3
cos(40 - x) + cos(40 - x) = sqrt 3
cos(40 - x) = sqrt 3/2
40 - x = arc cos(sqrt3/2) = 30
so x = 40 - 30 = 10 degree.
(3)
X is a point on AC such that BX is perpendicular to AC.
BX/BC = sin 45
BX = BC sin 45 = 2 x sqrt 2/2 = sqrt 2.
so area of triangle = (AC x BX)/2 = (2 sqrt 2)/2 = sqrt 2.

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