Rewrite the expression as an algebraic expresion in x:

tan(sin−1x)=

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
= x/√(1-x^2)

check:

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.

sorry I meant tan(sin^-1 x)=

Let's begin by breaking down the expression:

1. sin^(-1)(x) represents the inverse sine of x, which is commonly denoted as arcsin(x).
2. The tangent function of an angle is given by tan(theta).

So, the expression tan(sin^(-1)(x)) can be rewritten as:

tan(arcsin(x))

Now, let's convert it to an algebraic expression.

To rewrite the expression, we need to understand the inverse trigonometric function. The expression sin^(-1)x, also written as arcsin(x) or asin(x), represents the angle whose sine is x.

So, let's break it down step by step:

1. Let α be the angle such that sin α = x.

2. Taking the tan of both sides, we get:
tan(sin α) = tan(x).

Now, we know that tan(sin α) can be simplified using trigonometric identities. One identity that is useful in this case is:
tan(sin α) = x / √(1 - x^2).

Therefore, we can rewrite the expression as an algebraic expression in x:

tan(sin^(-1)x) = x / √(1 - x^2).

And that's the algebraic expression representing the given expression.