Prove that cot²(π/7)+cot²(3π/7)+cot²(5π/7) = 5?

pf :
cot 2α = ( cot²α - 1 )/( 2*cot α )
cot²α - 1 = 2( cot α )( cot 2α )
cot²α = 1 + 2( cot α )( cot 2α )

Let θ = π/7
cot²(π/7) = 1 + 2( cot π/7 )( cot 2π/7 ) = 1 + 2*cot θ*cot 2θ
cot²(3π/7) = 1 + 2( cot 3π/7 )( cot 6π/7 ) = 1 + 2( cot 3π/7 )( - cot π/7 ) = 1 - 2*cot θ*cot 3θ
cot²(5π/7) = 1 + 2( cot 5π/7 )( cot 10π/7 ) = 1 + 2( - cot 2π/7 )( cot 3π/7 ) = 1 - 2*cot 2θ*cot 3θ

cot²(π/7) + cot²(3π/7) + cot²(5π/7)
= 1 + 2*cot θ*cot 2θ + 1 - 2*cot θ*cot 3θ + 1 - 2*cot 2θ*cot 3θ
= 3 + 2( cot θ*cot 2θ - cot θ*cot 3θ - cot 2θ*cot 3θ )
= 3 + 2*1 , see Note below
= 5

Q.E.D.

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Note.
claim : cot θ*cot 2θ - cot θ*cot 3θ - cot 2θ*cot 3θ = 1
pf :
cot (α+β) = ( cot α*cot β - 1 ) / ( cot α + cot β )
Take α = θ , β = 2θ , then we have
cot 3θ = ( cot θ*cot 2θ - 1 ) / ( cot θ + cot 2θ )
cot θ*cot 2θ - 1 = cot 3θ*( cot θ + cot 2θ )
cot θ*cot 2θ - 1 = cot θ*cot 3θ + cot 2θ*cot 3θ
cot θ*cot 2θ - cot θ*cot 3θ - cot 2θ*cot 3θ = 1
Q.E.D.