All we need are the following inequalities: \[ \begin{aligned} \cot 47^{\circ} < 1 < \tan 47^{\circ} \\ \cos 47^{\circ} < \frac{\sqrt{2}}{2} < \sin 47^{\circ} \end{aligned} \] Also note that $ \sin 47^{\circ} $ and $ \cos 47^{\circ} $ are very close to $ \frac{\sqrt{2}}{2} $ and that $ \tan 47^{\circ} $ and $ \cot 47^{\circ} $ are very close to $ 1 $. Thus, applying those approximate values, we have the following: $ \textbf{(A)} $: $ \tan 47^{\circ} + \cos 47^{\circ} > \tan 47^{\circ} + \cos 45^{\circ} \approx 1 + \frac{\sqrt{2}}{2} $. $ \textbf{(B)} $ $ \cot 47^{\circ} + \sqrt{2}\sin 47^{\circ} > \cot 45^{\circ} + \sqrt{2}\sin 47^{\circ} \approx 1 + (\sqrt{2})(\frac{\sqrt{2}}{2}) \approx 1 + 1 \approx 2 $. $ \textbf{(C)} $ $ \sqrt{2}\cos 47^{\circ} + \sin 47^{\circ} > \sqrt{2}\cos 45^{\circ} + \sin 47^{\circ} \approx (\sqrt{2})(\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2} \approx 1 + \frac{\sqrt{2}}{2} $. $ \textbf{(D)} $ $ \tan 47^{\circ} + \cot 47^{\circ} \approx 1 + 1 \approx 2 $. $ \textbf{(E)} $ $ \cos 47^{\circ} + \sqrt{2}\sin 47^{\circ} \approx \frac{\sqrt{2}}{2} + (\sqrt{2})(\frac{\sqrt{2}}{2}) \approx 1 + \frac{\sqrt{2}}{2} $. Thus, we need only to compare $ \textbf{(B)} $ and $ \textbf{(D)} $. We have \[ \begin{aligned} \cot 47^{\circ} + \sqrt{2}\sin 47^{\circ} & \qquad \text{?} \qquad \tan 47^{\circ} + \cot 47^{\circ} \\ \implies \sqrt{2}\sin 47^{\circ} & \qquad \text{?} \qquad \tan 47^{\circ} \\ \implies \sqrt{2} & \qquad \text{?} \qquad \frac{1}{\cos 47^{\circ}} \\ \implies \cos 47^{\circ} & \qquad \text{?} \qquad \frac{\sqrt{2}}{2} \end{aligned} \] But we know that $ \cos 47^{\circ} < \frac{\sqrt{2}}{2} $. Thus our $ ? $ sin is just in fact a $ < $ sign and we have $ \cot 47^{\circ} + \sqrt{2}\sin 47^{\circ} < \tan 47^{\circ} + \cot 47^{\circ} $ and our answer is $ \boxed{\textbf{(D)} \tan 47^{\circ} + \cot 47^{\circ} } $.