急! F4 Basic Trigonometry 13q1

12a) (cos θ + 1)(cos θ - 1) = cos2 θ - 1 = - sin 2 θ

(sin θ + 1)(sin θ - 1) = sin2 θ - 1 = - cos2 θ

Hence

(cos θ + 1)(cos θ - 1)/[(sin θ + 1)(sin θ - 1)] = tan2 θ = 144/25

b) tan θ < 0, possible combinations are:

(i) sin θ = -12/13, cos θ = 5/13
(ii) sin θ = 12/13, cos θ = -5/13

For (i):

(cos θ + 1)(cos θ - 1)/[(sin θ + 1)(sin θ - 1)] = (5/13 + 1)(5/13 - 1)/[(-12/13 + 1)(-12/13 - 1)] = 144/25

For (ii):

(cos θ + 1)(cos θ - 1)/[(sin θ + 1)(sin θ - 1)] = (-5/13 + 1)(-5/13 - 1)/[(12/13 + 1)(12/13 - 1)] = 144/25

c) (a) is quicker.

20) sin2 10 + sin2 20 + sin2 30 + sin2 40 + sin2 50 + sin2 60 + sin2 70 + sin2 80

= sin2 10 + sin2 20 + sin2 30 + sin2 40 + cos2 40 + cos2 30 + cos2 20 + cos2 10

= 4

21) tan 100 tan 190 = tan (180 - 80) tan (180 + 10) = tan 80 tan 10 = (1/tan 10) tan 10 = 1

Similarly:

tan 110 tan 200 = 1

tan 120 tan 210 = 1

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tan 170 tan 260 = 1

Hence hen sum is 8.