`tan(sin^-1(x)) = x/sqrt(1 - x^2)` Verfiy the identity.

Textbook Question

Chapter 5, 5.2 - Problem 47 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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For both sides to have sense, x must be in [-1, 1]. For those x'es `sin^(-1)(x)` is the angle y in `[-pi/2, pi/2]` such that sin(y)=x.

cos is nonnegative on `[-pi/2, pi/2],` so `cos(y) = sqrt(1-sin^2(y)) = sqrt(1-x^2).` And

`tan(sin^(-1)(x)) = tan(y) = sin(y)/cos(y) = x/sqrt(1-x^2),` QED.

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balajia | College Teacher | (Level 1) eNoter

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Taking x as opposite side and 1 as hypotenuse we get adjacent side as `sqrt(1-x^2).`

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

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