Write an algebraic expression for tan(arccos x/2)

Let [arc cos(x/3)] = y 
Therefore, cos y = (x/3) now you have to express tany in terms of cosy: tany = siny/cosy = [±√(1 - cos²y)]/cosy note that, due to arccosine function range [0, π] and thus y itself ranging from 0 to π, it ends either in the 1st or i the 2nd quadrant and siny is then positive; therefore you can rewrite the previous expression taking the plus sign: tany = siny/cosy = [√(1 - cos²y)]/cosy hence, being cosy = (x/3), you get: 
tany = {√[1 - (x/3)²]} /(x/3) = {√[1 - (x²/9)]} /(x/3) = {√[(9 - x²)/9]} /(x/3) = (1/3){√[(9 - x²)} (3/x) = {√[(9 - x²)} / x therefore: 
tan [arccos(x/3)] = tany = {√[(9 - x²)} / x 

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