Determine the smallest positive value of $x$ (in degrees) for which $tan (x + 100^{\circ}) = tan (x + 50^{\circ}) tan x tan (x - 50^{\circ}) .$

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Solution

$tan (x + 100) = tan (x - 50) + tan x + tan (x + 50) \frac{tan (x + 100)}{tan (x - 50)} = tan x tan (x + 50) \frac{sin (x + 100) cos (x - 50)}{sin (x - 50) cos (x + 100)} = \frac{sin x cos (x + 50)}{sin (x + 50) cos x} 2 sin (2 x + 50) = \frac{- cos 50}{cos (2 x + 50)} sin (4 x + 100) = sin (270 + 50) 4 x + 100 = 320 x = 55$

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