le題數點計(maths)

3sinθ-2cosθ
——————
2cosθ

=(3sinθ / 2cosθ) - (2cosθ / 2cosθ)

=(3/2)*tanθ - 1

=(3/2)*4 - 1

=6 - 1

=5

For a right-angle triangle, angle A is 90deg, side a, b, c have relationship of
a^2 = b^2 + c^2
tan(B)= b/c = 4
sin(B)= b/a
cos(B)= c/a

(3sinB - 2cosB)/(2cosB)=(3*b/a -2*c/a)/(2*c/a)
=(3b-2c)/2c
=1.5*(b/c)-1
=1.5*4-1
=5