Find the value of:
$\tan {{43}^{\circ }}\tan {{60}^{\circ }}\tan {{47}^{\circ }}$

Hint: In the expression given above we can write angle ${{43}^{\circ }}$ as ${{90}^{\circ }}-{{47}^{\circ }}$ and then substitute this value of ${{43}^{\circ }}$ in $\tan {{43}^{\circ }}$ then you will get $\tan \left( {{90}^{\circ }}-{{47}^{\circ }} \right)$. Now, we know that $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ using this relation in $\tan \left( {{90}^{\circ }}-{{47}^{\circ }} \right)$ then you will find that $\tan {{43}^{\circ }}\And \tan {{47}^{\circ }}$ are complementary to each other. Then substitute the value of $\tan {{60}^{\circ }}$ in the given expression which is equal to $\sqrt{3}$. And hence, solve the expression.

Complete step-by-step answer:
We have to find the value of the following expression:
$\tan {{43}^{\circ }}\tan {{60}^{\circ }}\tan {{47}^{\circ }}$
In the above expression we can write angle ${{43}^{\circ }}$ as ${{90}^{\circ }}-{{47}^{\circ }}$ in $\tan {{43}^{\circ }}$.
$\tan \left( {{90}^{\circ }}-{{47}^{\circ }} \right)\tan {{60}^{\circ }}\tan {{47}^{\circ }}$
We know that:
$\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $
So, we can write $\tan \left( {{90}^{\circ }}-{{47}^{\circ }} \right)$ as $\cot {{47}^{\circ }}$ in the above expression so after substituting this value the expression will look like:
$\cot {{47}^{\circ }}\tan {{60}^{\circ }}\tan {{47}^{\circ }}$
We know from the trigonometric ratios that:
$\cot \theta =\dfrac{1}{\tan \theta }$
So, we can write $\cot {{47}^{\circ }}$ as $\dfrac{1}{\tan {{47}^{\circ }}}$ in the above expression.
$\dfrac{1}{\tan {{47}^{\circ }}}\left( \tan {{60}^{\circ }}\tan {{47}^{\circ }} \right)$
In the above expression you can see that $\tan {{47}^{\circ }}$ will be cancelled out in the numerator and denominator.
$\tan {{60}^{\circ }}$
From the trigonometric ratios we know the value of $\tan {{60}^{\circ }}=\sqrt{3}$ so substituting this value in the above expression we get,
$\sqrt{3}$
From the above solution we have got the value of the given expression as $\sqrt{3}$.

Note: In the above solution, instead of writing angle ${{43}^{\circ }}$ as ${{90}^{\circ }}-{{47}^{\circ }}$ in $\tan {{43}^{\circ }}$ we can write the angle ${{47}^{\circ }}$ as ${{90}^{\circ }}-{{43}^{\circ }}$ in $\tan {{47}^{\circ }}$ then the given expression will look like:
$\tan {{43}^{\circ }}\tan {{60}^{\circ }}\tan \left( {{90}^{\circ }}-{{43}^{\circ }} \right)$
Now, we can write $\tan \left( {{90}^{\circ }}-{{43}^{\circ }} \right)$ as $\cot {{43}^{\circ }}$ in the above expression.
$\tan {{43}^{\circ }}\tan {{60}^{\circ }}\cot {{43}^{\circ }}$
We can also use the relation between $\tan \theta \And \cot \theta $ in the above expression which is equal to:
$\cot \theta =\dfrac{1}{\tan \theta }$
$\tan {{43}^{\circ }}\tan {{60}^{\circ }}\left( \dfrac{1}{\tan {{43}^{\circ }}} \right)$
In the above expression, $\tan {{43}^{\circ }}$ will be cancelled out and we get,
$\tan {{60}^{\circ }}$
In the above solution part, we have shown that $\tan {{60}^{\circ }}=\sqrt{3}$ so using this relation we have got the above expression equivalent to:
$\sqrt{3}$
As you can see that we are getting the same as that we were getting in the solution part so this method is also correct.