已知幂函数$f\left( x \right)={{x}^{\alpha }}$的图象经过点$\left( 2,\dfrac{1}{4} \right)$，若$a=f\left( \sin\dfrac{4\pi}{7} \right)$，$b=f\left( \cos\dfrac{4\pi}{7} \right)$，$c=f\left( \tan\dfrac{4\pi}{7} \right)$，则$(\qquad)$．

$c\\lt a\\lt b$

$b\\lt c\\lt a$

$a\\lt b\\lt c$

$a\\lt c\\lt b$

$\because $ 幂函数$f\left( x \right)={{x}^{\alpha }}$的图象经过点$\left( 2,\dfrac{1}{4} \right)$，

$\therefore {{2}^{\alpha }}=\dfrac{1}{4}$，

$\therefore \alpha=-2$，

$\therefore f\left( x \right)={{x}^{-2}}$，

其定义域为$\left( -\infty ,0 \right)\cup \left( 0,+\infty \right)$关于原点对称，又$f\left( -x \right)={{\left( -x \right)}^{-2}}={{x}^{-2}}=f\left( x \right)$，

$\therefore f\left( x \right)={{x}^{-2}}$为偶函数，且在$\left( 0,+\infty \right)$上单调递减，

$a=f\left( \sin\dfrac{4\pi}{7} \right)=f\left( \sin\dfrac{3\pi}{7} \right)$，$b=f\left( \cos\dfrac{4\pi}{7} \right)=f\left( -\cos\dfrac{3\pi}{7} \right)=f\left( \cos\dfrac{3\pi}{7} \right)=f\left( \sin\dfrac{\pi}{14} \right)$，

$c=f\left( \tan\dfrac{4\pi}{7} \right)=f\left( -\tan\dfrac{3\pi}{7} \right)=f\left( \tan\dfrac{3\pi}{7} \right)$，

$\because y=\sin x$在$\left( 0,\dfrac{\pi}{2} \right)$内单调递增，则$\sin\dfrac{\pi}{14}\lt \sin\dfrac{\text{3 }\!\!\pi}{7}$，

又$\because 0\lt \sin\dfrac{\text{3 }\!\!\pi}{7}\lt 1$，$0\lt \cos \dfrac{\text{3 }\!\!\pi}{7}\lt 1$，则$\tan\dfrac{\text{3 }\!\!\pi}{7}=\dfrac{\sin \dfrac{\text{3 }\!\!\pi}{7}}{\cos \dfrac{\text{3 }\!\!\pi}{7}}\gt \sin \dfrac{\text{3 }\!\!\pi}{7}$，

$\therefore \tan\dfrac{\text{3 }\!\!\pi}{7}\gt \sin \dfrac{\text{3 }\!\!\pi}{7}\gt \sin \dfrac{\pi}{14}\gt 0$，

$\therefore f\left( \tan\dfrac{3\pi}{7} \right)\lt f\left( \sin\dfrac{3\pi}{7} \right)\lt f\left( \sin\dfrac{\pi}{14} \right)$，

$\therefore f\left( \tan\dfrac{4\pi}{7} \right)\lt f\left( \sin\dfrac{4\pi}{7} \right)\lt f\left( \cos\dfrac{\text{4 }\!\!\pi}{7} \right)$，即$c\lt a\lt b$．

故选：$\rm A$