Simplify the appearance.

tan [sin^-1 (x)]

allow u = sin^-1 (x)

therefore, sin u = x

today u draw a right-angled triangle with a perspective u :

from sin u = x , the opp. is x together with hypotenus is 1, after that u discover the adj. that is √(1 – x^2)

tan [sin^-1 (x)]

= tan u

= opp./adj.

= x / √(1 – x^2)

consequently

tan [sin^-1 (x)]

= x / √(1 – x^2)

Simplify This Phrase

If by sin^-1(x) you imply the inverse purpose of sin after that continue usually disregard this this solution.

OK therefore genuinely believe that sin^-1(x) = T after that by making use of the sin purpose to both edges we now have

sin(sin^-1(x)) = sin(T)

x = sin(T) therefore, sin(T) = x

today since sin(T) = opposite/hypotenuse = x

therefore allow the hypotenuse = 1 and so the other part of T = x

therefore we need certainly to discover the dimension regarding the part right beside T with the Pythagorean theorem we have

the square-root of (1 – x^2)

today since tan = opposite/adjacent

we now have

tan(sin^-1(x)) = tan(T) = x / [root(1-x^2) ]

draw the right angled triangle, with hypotenuse = 1, side=x —>other part = sqrt(1-x^2)

Therefore tan(sin^-1(x)) = x/sqrt(1-x^2)