`tan(sin^-1((x - 1)/4)) = (x -1 )/sqrt(16 - (x-1)^2)` Verfiy the identity.

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Chapter 5, 5.2 - Problem 49 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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For `sin^(-1)((x-1)/4)` to have sense, (x-1)/4 must be in [-1, 1]. In this case, `sin^(-1)((x-1)/4)` is an angle y in `[-pi/2, pi/2]` such that sin(y) = (x-1)/4.

For those angles `cos(y) gt= 0` and

`cos(y) = +sqrt(1-sin^2(y)) = sqrt(1-((x-1)/4)^2) = (1/4)*sqrt(16-(x-1)^2).`

`tan(sin^(-1)((x-1)/4)) = tan(y) = sin(y)/cos(y) = ((x-1)/4)/(1/4*sqrt(16-(x-1)^2)) = (x-1)/sqrt(16-(x-1)^2),` QED.

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balajia | College Teacher | (Level 1) eNoter

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Taking (x-1) as opposite side and 4 as hypotenuse,we get adjacent side as `sqrt(4^2-(x-1)^2)` .

`tan(sin^-1((x-1)/4))`  = `tan(tan^-1((x-1)/sqrt(16-(x-1)^2)))` = `(x-1)/sqrt(16-(x-1)^2).`

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