Answer
$sin(arctan\frac{3}{4})=\frac{3}{5}$
Work Step by Step
Let $u=arctan\frac{3}{4}$. Then:
$tan~u=\frac{3}{4}$
$cot~u=\frac{1}{tan~u}=\frac{4}{3}$
The range $arctan~x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $arctan\frac{3}{4}\gt0$, then $0\lt u\lt\frac{\pi}{2}$ (First Quadrant)
$csc^2u=1+cot^2u$
$csc^2u=1+(\frac{4}{3})^2$
$csc^2u=1+\frac{16}{9}=\frac{25}{9}$
$csc~u=\frac{5}{3}$ (First Quadrant)
$sin(arctan\frac{3}{4})=sin~u=\frac{1}{csc~u}=\frac{3}{5}$
