Precalculus, Chapter 5, 5.2, Section 5.2, Problem 50

You need to rremember that tan^2 theta+ 1 = 1/(cos^2 theta) , hence, replacing cos^(-1) ((x+1)/2) by alpha , yields:
cos^(-1) ((x+1)/2) = alpha => cos(cos^(-1) ((x+1)/2)) = cos alpha
(x+1)/2 = cos alpha
Raising to square both sides, yields:
(x+1)^2/4 = cos^2 alpha =>1/(cos^2 alpha) = 4/((x+1)^2)
But 1/(cos^2 alpha) = tan^2 alpha + 1 , hence 4/((x+1)^2) = tan^2 alpha + 1.
tan^2 alpha = 4/((x+1)^2) - 1
tan^2 alpha = (4 - (x+1)^2)/((x+1)^2)
You need to take square root both sides, such that:
tan^2 alpha = (4 - (x+1)^2)/((x+1)^2)
tan alpha = sqrt((4 - (x+1)^2)/((x+1)^2))
tan alpha = sqrt((4 - (x+1)^2)/((x+1)^2))
Hence, verifying the given identity tan(arccos((x+1)/2)) = (sqrt(4-(x+1)^2))/(x+1) yields that it is true.