Please answer the following question, thank you?

(a)

Use the hint:

sin²A = (1/2) (1 - cos 2A) ...[1]

cos²A = (1/2) (1 + cos 2A) ...[2]

[1]÷[2] gives

tan²A = (1 - cos 2A)/(1 + cos 2A)

Put A = π/4 - θ/2,

tan²(π/4 - θ/2)

= [1 - cos(π/2 - θ)]/[1 + cos(π/2 - θ)]

= (1 - sinθ)/(1 + sinθ)

= (cosθ - sinθcosθ)/(cosθ + sinθcosθ)

= (2cosθ - 2sinθcosθ)/(2cosθ + 2sinθcosθ)

= [2cosθ - sin(2θ)]/[2cosθ + sin(2θ)]

(b)

Note that for 0 < θ < π/2,

we have 0 < θ/2 < π/4,

and -π/4 < -θ/2 < 0,

i.e. 0 < π/4 - θ/2 < π/4.

Using the result of (a), the equation is

tan²(π/4 - θ/2) = 1/3

tan(π/4 - θ/2) = 1/√3 or -1/√3 (rejected for 0 < π/4 - θ/2 < π/4)

π/4 - θ/2 = π/6

θ/2 = π/4 - π/6

θ/2 = π/12

θ = π/6