posted by Anonymous .
Rewrite the expression as an algebraic expresion in x:
sorry I meant tan(sin^-1 x)=
sin^-1 x is angle in a right - angled triangle where the opposite side is x and the hypotenuse is 1
let that angle be Ø
so really sin^-1 x ----> sinØ = x/1
so the third side is √(1 - x^2)
then tan(sin^-1 x)
= tan Ø
e.g. let x = .5
then sin^-1 (.5) = π/6
and tan(π/6) = .57735
according to my answer
tan(sin^-1 (.5)) = .5/√(1-.25) = .57735
all looks good.