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Evaluate the following:
\[{\text{tan 3}}{{\text{5}}^ \circ }{\text{ tan 4}}{{\text{0}}^ \circ }{\text{ tan 4}}{{\text{5}}^ \circ }{\text{ tan 5}}{{\text{0}}^ \circ }{\text{ tan 5}}{{\text{5}}^ \circ }\]

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Hint: Use the conversion of tangent to cotangent.

To evaluate,
\[{\text{tan 3}}{{\text{5}}^ \circ }{\text{ tan 4}}{{\text{0}}^ \circ }{\text{ tan 4}}{{\text{5}}^ \circ }{\text{ tan 5}}{{\text{0}}^ \circ }{\text{ tan 5}}{{\text{5}}^ \circ }\]

We know that in trigonometry,
$\tan (90 - \theta ) = \cot \theta $
and $\cot \theta = \dfrac{1}{{\tan \theta }}$

So by using this identity we will get,
\[
  tan{35^ \circ } = tan({90^ \circ } - {55^ \circ }) = cot{55^ \circ } \\
  tan{40^ \circ } = tan({90^ \circ } - {50^ \circ }) = cot{50^ \circ } \\
\]
Now if we substitute these values in our original question we get,
\[
   = cot{55^ \circ }cot{50^ \circ }tan{45^ \circ }tan{50^ \circ }tan{55^ \circ } \\
   = \dfrac{1}{{tan{{55}^ \circ }}} \times \dfrac{1}{{tan{{50}^ \circ }}} \times tan{45^ \circ } \times tan{50^ \circ } \times tan{55^ \circ } \\
 \]

On solving it we get,
$\tan {45^ \circ }$ = $1$

Hence, the answer is $1$

Note: In these types of problems, the conversion from one trigonometric quantity to another is crucial. Also, it's helpful to remember the trigonometric values.

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