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**Hint:**As we can see that the above question is related to trigonometry as tangent and cotangent are trigonometric ratios. WE can solve this question by applying the trigonometric identities. Some of the basic identities are $\tan (90 - \theta ) = \cot \theta $ and we can write $\cot (90 - \phi ) = \tan \phi $. We should also know that $\cot \theta $ can be written as $\dfrac{1}{{\tan \theta }}$.

**Complete step by step answer:**

Here we have $\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}}$.

By applying the identities $\tan (90 - \theta ) = \cot \theta $ and $\cot (90 - \phi ) = \tan \phi $ in the denominator we can solve this. By comparing for tangent we have $\theta = 57$, and for cotangent we have $\phi = 37$.

So we can write $\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{(90 - 57)}^ \circ } + \cot {{(90 - 37)}^ \circ }}}$. It can be written as $\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\cot {{57}^ \circ } + \tan {{37}^ \circ }}}$.

We know that $\cot \phi $ can be written as $\dfrac{1}{{\tan \phi }}$ and the same for tan $\theta $.

We can write the expression as $\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\dfrac{1}{{\tan {{57}^ \circ }}} + \dfrac{1}{{\cot {{37}^ \circ }}}}}$.

By taking the LCM of the denominator of the denominator we can write $\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\dfrac{{\cot 37 + \tan 57}}{{\tan {{57}^ \circ } \cdot \cot {{37}^ \circ }}}}}$.

On further solving we can write $\tan {57^ \circ } \cdot \cot 37\left[ {\dfrac{{\tan 57 + \cot 37}}{{\tan 57 + \cot 37}}} \right]$.

So it gives us the value $\tan {57^ \circ } \cdot \cot {37^ \circ }$.

**Hence, the correct answer is option B.**

**Note:**We should note that $\tan \theta $ can also be written as $\dfrac{1}{{\cot \theta }}$. Before solving this kind of question we should have the full knowledge of trigonometric functions and their identities. We know that if there is $\dfrac{a}{{\dfrac{b}{c}}}$, then it can be written as $\dfrac{{c \times a}}{b}$. This is what we have applied in the above solution.

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