已知\sin \varphi = -\dfrac{\sqrt{3}}{2}，\mathrel{|} \varphi \mathrel{|} \lt \dfrac{\pi }{2}，
故有，-\dfrac{\pi }{2}\lt \varphi \lt 0，
故有\cos \varphi = \sqrt{1-\sin ^{2}\varphi }= \dfrac{1}{2}，
故有\tan \varphi = \dfrac{\sin \varphi }{\cos \varphi }= -\sqrt{3}，
故答案为：-\sqrt{3}．
