Question 612706
arcsin(x) represents the angle whose sin is x. So tan(arcsin(x)) represents the tan ratio for that angle.<br>
It will be easier to understand if you...<ol><li>Draw a right triangle.</li><li>Pick one of the acute angles (i.e not the right angle) to be the arcsin(x) angle. Let's call this angle A.</li><li>Since the sin ratio is opposite/hypotenuse and since we want the sin ration to be x, label the side opposite to A as "x" and the hypotenuse as "1". It should be clear that the sin ration for A is x/1 = x.<br>Since we want the tan ratio for angle A and since the tan ratio is opposite/adjacent, we need the adjacent side to angle A. Let's label the adjacent side "y".</li><li>Use the Pythagorean theorem  and solve for y:
{{{x^2 + y^2 = 1^2}}}
{{{x^2 + y^2 = 1}}}
{{{y^2 = 1-x^2}}}
{{{y = sqrt(1-x^2)}}}</li><li>Write the tan ratio for A, using the square root expression for the adjacent side:
tan(arccos(x)) = {{{x/sqrt(1-x^2)}}}
This may be an acceptable solution to your problem. But...</li><li>Our expression for tan(arccos(x)) has a square root in its denominator. Usually these denominators get rationalized:
tan(arccos(x)) = {{{x/sqrt(1-x^2) = (x/sqrt(1-x^2))(sqrt(1-x^2)/sqrt(1-x^2)) = (x*sqrt(1-x^2))/(1-x^2)}}}</li></ol>