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If α, β and γ are the eccentric angles of three points on the ellipse x2/a2 + y2/b2 = 1 at which the normals are concurrent, then show that sin(α + β) + sin(β + γ) + sin(γ + α) = 0

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Suppose the normals at α, β and  γ are concurrent at (h, k) and let δ be the foot of the fourth normal from (h, k) Then we have

∑tan α/2 tan β/2 = 0

and tan α/2 tan β/2 tan γ/2 tan δ/2 = -1

Eliminating tan δ/2 from the above two equations gives

0 = ∑tan (α/2) tan (β/2)

Therefore

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