Find sin(tan^(-1)(x+1)).

Hello!
The domain of this function is all real numbers, because tan^(-1)(x), and therefore tan^(-1)(x+1), is defined for all x in RR.
By the definition, tan^(-1)(y) is the number w in (-pi/2, pi/2) such that y = tan(w). Knowing tan(w), we can find sin(w) = sin(tan^(-1)(y)),
namely
(tan^2(w))/(1+tan^2(w)) = ((sin^2(w))/(cos^2(w))) / (1+(sin^2(w))/(cos^2(w))) = sin^2(w).
This way sin(w) = +-(tan(w))/sqrt(1+tan^2(w)) (= +-y/sqrt(1+y^2)).
Note that for w in (-pi/2, pi/2)  cos(w) gt 0, hence the signs of sin(w) and tan(w) are the same. The formula becomes  sin(w) = (tan(w))/sqrt(1+tan^2(w)).
So we obtained that  sin(tan^(-1)(y)) = y/sqrt(1+y^2).
Thus the answer for our question is (substitute y = x+1 )
sin(tan^(-1)(x+1)) = (x+1)/sqrt(1+(x+1)^2)
for all x in RR.
 

Imagine a right angled triangle with angle theta at one vertex and opposite side (x+1) and adjacent side equal to 1. tan theta = (x+1)/1. So theta = tan^-1(x+1).
By Pythagoras theorem, the length of the hypotenuse is sqrt{(x+1)^2 +1}=sqrt(x^2 +2x + 2)
so sin theta = sin( tan^-1(x+1) = opposite/hypotenuse = (x+1)/(x^2+2x+2)