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