The trig ratios for these angles are all fairly easy to work out. First 45 degrees: a right-angled isosceles triangle has two angles of 45 degrees. The hypotenuse is the sum of the squares of two equal sides. If we make the sides containing the right angle length 1, then the ratio of the either side to the hypotenuse is the same. In other words the sine and cosine of 45 is 1/sqrt(1+1)=1/sqrt(2). The tangent of 45 is one because tan=sin/cos and because sin=cos tan45=1.
In an equilateral triangle all angles are 60 degrees, and if we drop a perpendicular from a vertex on to the opposite side we have two congruent right-angled triangle back to back. The perpendicular bisects the angle and the side, so the other two angles of the right-angled triangles are 30 and 60 degrees. We know the lengths of these sides because we started with an equilateral triangle. If we make the sides length 1, then the bisected side is length 1/2. The hypotenuses of the right-angled triangle are both 1 and the side opposite the 30 degree angle is length 1/2. What is the length of the perpendicular? Its square=1-(1/2)^2=1-1/4=3/4, so the length is sqrt(3)/2. The sine of 30=(opposite)/(hypotenuse)=1/2. This is also the cosine of 60 because the opposite side to 30 is the adjacent side to 60. The cosine of 30=sine of 60=sqrt(3)/2 and the tangent of 60 is sqrt(3)/2 divided by 1/2=sqrt(3). The tangent of 30 is the reciprocal of tan60 because opposite and adjacent swap places, so tan30=1/sqrt(3).
Returning to the question we have:
2(cos^2(45)+tan^2(60))-6(sin^2(45)+tan^2(30))=2(1/2+3)-6(1/2+1/3)=2(7/2)-6(5/6)=7-5=2, not 6 as the question indicates.