A geometry problem by Vaibhav Prasad

$\cot{\dfrac{4\pi}{7}}\cot{\dfrac{8\pi}{7}} - \cot{\dfrac{8\pi}{7}}\cot{\dfrac{12\pi}{7}} - \cot{\dfrac{12\pi}{7}}\cot{\dfrac{4\pi}{7}} \\ = \cot{\dfrac{4\pi}{7}} \cot{\dfrac{8\pi}{7}} \cot{\dfrac{12\pi}{7}} \left( \tan{\dfrac{12\pi}{7}} - \tan{\dfrac{4\pi}{7}} - \tan{\dfrac{8\pi}{7}} \right) \\ = \dfrac {\tan{\color{#D61F06}{\dfrac{12\pi}{7}}} - \tan{\dfrac{4\pi}{7}} - \tan{\color{#3D99F6}{\dfrac{8\pi}{7}}}} {\tan{\dfrac{4\pi}{7}} \tan{\color{#3D99F6}{\dfrac{8\pi}{7}}} \tan{\color{#D61F06}{\dfrac{12\pi}{7}}}} \\ = \dfrac {\color{#D61F06}{-}\tan{\color{#D61F06}{\dfrac{2\pi}{7}}} - \tan{\dfrac{4\pi}{7}} - \tan{\color{#3D99F6}{\dfrac{\pi}{7}}}} {\color{#D61F06}{-} \tan{\dfrac{4\pi}{7}} \tan{\color{#3D99F6}{\dfrac{\pi}{7}}} \tan{\color{#D61F06}{\dfrac{2\pi}{7}}}} \\ = \dfrac {\tan{\dfrac{\pi}{7}} + \tan{\dfrac{2\pi}{7}} + \tan{\dfrac{4\pi}{7}}} {\tan{\dfrac{\pi}{7}} \tan{\dfrac{2\pi}{7}} \tan{\dfrac{4\pi}{7}}} = \boxed{1} \quad \quad \color{#D61F06} {\text{See Note}}$

$$

$\color{#D61F06} {\text{Note: }}$ Since $\frac{\pi}{7} + \frac{2\pi}{7} + \frac{4\pi}{7} = \pi$ , angles of a triangle. $\Rightarrow \tan{\frac{\pi}{7}} + \tan{\frac{2\pi}{7}} + \tan{\frac{4\pi}{7}} = \tan{\frac{\pi}{7}} \tan{\frac{2\pi}{7}} \tan{\frac{4\pi}{7}}$

α + β + γ = 3π/2

tan(α + β) = tan(3π/2 - γ)

(tanα + tanβ)/(1 - tanαtanβ ) = 1/tanγ

tanαtanβ + tanβtanγ + tanγtanα = 1

cot(4π/7)cot(8π/7)-cot(8π/7)cot(12π/7)-cot(12π/7)cot(4π/7)

= tan(13π/14)tan(5π/14)+tan(5π/14)tan(3π/14)+tan(3π/14)tan(13π/7)

= 1

13π/14 + 5π/14 + 3π/14 = 3π/2