Answer
The identity is verified.
$tan^{-1}(sin\frac{x-1}{4})=\frac{x-1}{\sqrt {16-(x-1)^2}}$
Work Step by Step
Make $z=\frac{x-1}{4}$
$y=sin^{-1}z$, domain: $-1\leq z\leq1$, range: $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$:
$z=sin~y$
$z^2=sin^2y$
$1-z^2=1-sin^2y$
$1-z^2=cos^2y$
$cos~y=+\sqrt {1-z^2}~~$ (Use only the $+$ because $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$)
$tan^{-1}(sin\frac{x-1}{4})=tan(sin^{-1}z)=tan~y=\frac{sin~y}{cos~y}=\frac{z}{\sqrt {1-z^2}}=\frac{\frac{x-1}{4}}{\sqrt {1-(\frac{x-1}{4})^2}}=\frac{\frac{x-1}{4}}{\sqrt {1-\frac{(x-1)^2}{16}}}\frac{4}{4}=\frac{x-1}{\sqrt {16-(x-1)^2}}$
