Write the expression sin (tan^-1) x) as an algebraic expression in x (without trig or inverse trig functions).

A. (sqrt (1+x^2)) / (x)

B. (1) / (sqrt (1+x^2))

C. (x) / (sqrt (1-x^2))

D. (x) / (sqrt (1+x^2))

E. (1) / (sqrt (1-x^2))

Option "D" (x) / (sqrt (1+x^2)) is correct.

As sin^2 (α) + cos^2 (α) = 1

By division of sin^2 (α) on both sides of equation, we get

1 + 1/tan^2 (α) = 1/sin^2 (α)

By taking L. C. M and inverting the whole equation;

Sin^2 (α) = tan^2 (α) / (tan^2 (α) + 1)

Let α = arctan (x);

Sin^2 (arctan (x)) = tan^2 (arctan (x)) / (tan^2 (arctan (x)) + 1)

As tan (arctan (x)) = x

Hence

Sin^2 (arctan (x)) = x^2 / (x^2 + 1)

Now taking square root;

Sin (arctan (x)) = (x) / (sqrt (1+x^2))