Algebra and Trigonometry 10th Edition

The identity is veriified. $tan(cos^{-1}\frac{x+1}{2})=\frac{\sqrt {4-(x+1)^2}}{x+1}$
Make $z=\frac{x+1}{2}$ $y=cos^{-1}z$, domain: $-1\leq z\leq1$, range: $0\leq y\leq\pi$: $z=cos~y$ $z^2=cos^2y$ $1-z^2=1-cos^2y$ $1-z^2=sin^2y$ $sin~y=+\sqrt {1-z^2}~~$ (Use only the $+$ because $0\leq y\leq\pi$) $tan(cos^{-1}\frac{x+1}{2})=tan(cos^{-1}z)=tan~y=\frac{sin~y}{cos~y}=\frac{\sqrt {1-z^2}}{z}=\frac{\sqrt {1-(\frac{x+1}{2})^2}}{\frac{x+1}{2}}=\frac{\sqrt {1-\frac{(x+1)^2}{4}}}{\frac{x+1}{2}}\frac{2}{2}=\frac{\sqrt {4-(x+1)^2}}{x+1}$