#### Answer

$\dfrac {\pm \sqrt {2x+1}}{x+1}$

#### Work Step by Step

$\sin \theta =\pm \sqrt {1-\cos ^{2}\theta }=\pm \sqrt {1-\left( \dfrac {x}{x+1}\right) ^{2}}=\pm \sqrt {\dfrac {\left( x+1\right) ^{2}-x^{2}}{\left( x+1\right) ^{2}}}=\pm \dfrac {\sqrt {\left( x+1-x\right) \left( x+1+x\right) }}{\left( x+1\right) }=\dfrac {\pm \sqrt {2x+1}}{x+1}$