Answer
$\dfrac{\sqrt{x^2+2x-8}}{x+1}$
Work Step by Step
$\theta = \sec^{-1} \dfrac{x+1}{3}$
$\sec{\theta} = \dfrac{x+1}{3}$
$\cos{\theta} = \dfrac{3}{x+1}$
$\sin^2{\theta} = 1-\cos^2{\theta} = 1-\dfrac{9}{(x+1)^2}$
$\sin^2{\theta} = \dfrac{(x+1)^2-9}{(x+1)^2} = \dfrac{x^2+2x-8}{(x+1)^2}$
$\sin{\theta} = \dfrac{\sqrt{x^2+2x-8}}{x+1}$