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.`