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

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### 2 Answers

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.

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).`