What is the easiest way to show that (see picture)?

π/96 = (π/6)/16 so one idea is to cascade tan(θ/2) results

Let T = tan(2θ) and t = tan(θ)

T = 2t/(1 – t^2) and the lesser root for t is

t = [√(T^2 + 1) – 1]/T or if more convenient

t = √(1 + 1/T^2) – 1/T ..........................(1)

Start with T = tan(π/6) = √(3)/3 so that 1/T = √(3)

t = tan(π/12) = √(1 + 3) – √(3) = 2 – √(3)

Repeat that, but now T = tan(π/12) and 1/T = 2 + √(3), 1/T^2 = 7 + 4√(3)

tan(π/24) = t = √[1 + 1/T^2] – 1/T = √[8 + 4√(3)] – 2 - √(3)

Now you can see a plan to get to tan(π/48) and tan(π/96)

Repeating again, but with T = tan(π/24) = √[8 + 4√(3)] – 2 - √(3)

Expand T^2 and then note √(T^2) to keep to keep results in surd form.

But now it is your turn

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