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+
c. (x) / (sqrt (1-
d. (x) / (sqrt (1+
e. (1) / (sqrt (1-x^2))

The correct answer for this question is letter "D. (x) / (sqrt (1+x^2))."

Make a triangle. yes a triangle. label "opposite" side x and adjacent side 1 since the tangent is x = x/1
use pythatoras to find the hypotenuse which will be

square root (1 + x^2)

then
sin(tan−1(x)) = x / (square root (1 + x^2))

that being opposite over hypotenuse
and by the same triangle

cos(tan−1(x)) = x / (square root (1 + x^2))

(6,0) i hope it makes enough sense