Simplify the expression. tan(sin^-1(x))?

tan [sin^-1 (x)]

let u = sin^-1 (x)

so, sin u = x

now u draw a right-angled triangle with an angle u :

from sin u = x , the opp. is x and the hypotenus is 1, then u find the adj. which is √(1 - x^2)

tan [sin^-1 (x)]

= tan u

= opp./adj.

= x / √(1 - x^2)

therefore

tan [sin^-1 (x)]

= x / √(1 - x^2)

We want to find tan(sin^-1(x))

let arcsin(x) = u

then, x = sin(u) and we really just need to find out what tan(u) is.

sin²(u) + cos²(u) = 1

x² + cos²(u) = 1

cos²(u) = 1-x²

cos(u) = √(1-x²)

tan(u) = sin(u)/cos(u)

= x / √(1-x²)