设$a=\ln 1.1$，$b={{\text{e}}^{0.1}}-1$，$c=\tan 0.1$，$d=\dfrac{0.4}{\pi }$，则$(\qquad)$．

$a\\lt b\\lt c\\lt d$

$a\\lt c\\lt b\\lt d$

$a\\lt b\\lt d\\lt c$

$a\\lt c\\lt d\\lt b$

设$a\left( x \right)=\ln \left( x+1 \right)$，$b\left( x \right)={{\text{e}}^{x}}-1$，$c\left( x \right)=\tan x$，$d\left( x \right)=\dfrac{4}{\pi }x$，易得$a\left( 0 \right)=b\left( 0 \right)=c\left( 0 \right)=d\left( 0 \right)$$.$

设$y=d\left( x \right)-b\left( x \right)=\dfrac{4}{\pi }x-{{\text{e}}^{x}}+1$，则令${y}'=\dfrac{4}{\pi }-{{\text{e}}^{x}}=0$有$x=\ln \dfrac{4}{\pi }$，故$y=d\left( x \right)-b\left( x \right)$在$\left( -\infty ,\ln \dfrac{4}{\pi } \right)$上单调递增$.$

①$\because {{\left( \dfrac{4}{\pi } \right)}^{10}}\gt {{\left( \dfrac{4}{3.2} \right)}^{10}}={{\left( \dfrac{5}{4} \right)}^{10}}={{\left( \dfrac{25}{16} \right)}^{5}}\gt {{\left( \dfrac{24}{16} \right)}^{5}}={{\left( \dfrac{3}{2} \right)}^{5}}\gt \text{e}$，即${{\left( \dfrac{4}{\pi } \right)}^{10}}\gt \text{e}$，故$10\ln \dfrac{4}{\pi }\gt 1$，即$\ln \dfrac{4}{\pi }\gt 0.1$，故$d\left( 0.1 \right)-b\left( 0.1 \right)\gt d\left( 0 \right)-b\left( 0 \right)=0$，即$d\gt b$

②设$y=b\left( x \right)-c\left( x \right)={{\text{e}}^{x}}-1-\tan x$，则${y}'={{\text{e}}^{x}}-\dfrac{1}{{{\cos }^{2}}x}=\dfrac{{{\text{e}}^{x}}{{\cos }^{2}}x-1}{{{\cos }^{2}}x}$，设$f\left( x \right)={{\text{e}}^{x}}{{\cos }^{2}}x-1$，则${f}'\left( x \right)={{\text{e}}^{x}}\left( {{\cos }^{2}}x-2\sin x \right)={{\text{e}}^{x}}\left( -{{\sin }^{2}}x-2\sin x+1 \right)$

设$g\left( x \right)=x-\sin x$，则${g}'\left( x \right)=1-\cos x\ge 0$，故$g\left( x \right)=x-\sin x$为增函数，故$g\left( x \right)\ge g\left( 0 \right)=0$，即$x\ge \sin x$

故${f}'\left( x \right)\ge {{\text{e}}^{x}}\left( -{{x}^{2}}-2x+1 \right)={{\text{e}}^{x}}\left[ -{{\left( x+1 \right)}^{2}}+2 \right]$，当$x\in \left[ 0,0.1 \right]$时${f}'\left( x \right)\gt 0$， $f\left( x \right)={{\text{e}}^{x}}{{\cos }^{2}}x-1$为增函数，故$f\left( x \right)\ge {{\text{e}}^{0}}{{\cos }^{2}}0-1=0$，故当$x\in \left[ 0,0.1 \right]$时$y=b\left( x \right)-c\left( x \right)$为增函数，故$b\left( 0.1 \right)-c\left( 0.1 \right)\gt b\left( 0 \right)-c\left( 0 \right)=0$，故$b\gt c$

③设$y=c\left( x \right)-a\left( x \right)=\tan x-\ln \left( x+1 \right)$，${y}'=\dfrac{1}{{{\cos }^{2}}x}-\dfrac{1}{x+1}=\dfrac{x+{{\sin }^{2}}x}{\left( x+1 \right){{\cos }^{2}}x}$，易得当$x\in \left( 0,0.1 \right)$时${y}'\gt 0$，故$c\left( 0.1 \right)-a\left( 0.1 \right)\gt c\left( 0 \right)-a\left( 0 \right)=0$，即$c\gt a$

综上$d\gt b\gt c\gt a$．

故选：$\rm B$