Prove the following using trigonometric identities: $\tan {{20}^{\circ }}\tan {{80}^{\circ }}\cot {{50}^{\circ }}=\sqrt{3}$ \[\]

Hint: We proceed from left hand side of the proof and express the left hand side as $\tan \left( {{50}^{\circ }}-{{30}^{\circ }} \right)\tan \left( 50+{{30}^{\circ }} \right)\cot {{50}^{\circ }}$. We use reciprocal relationship $\cot A=\dfrac{1}{\tan A}$ to convert $\cot {{50}^{\circ }}$to $\tan {{50}^{\circ }}$, tangent angle difference formula $\tan \left( A+B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ and tangent angle sum formula $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ to simplify. W use tangent triple angle formula to $\tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}$ to arrive at the right hand side. \[\]

Complete step by step answer:
We know from tangent sum of angles formula (also known as compound angle formula ) that for two angles with measure $A,B$ then the tangent of the their sum can be given as
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]
We tangent difference of angle formula that the tangent of difference of angles $A,B$with $A>B$ is given by
\[\tan \left( A+B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\]
We know from tangent triple angle formula that
\[\tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}\]
We also know the reciprocal relationship between tangent and cotangent of any angle $A$ as
\[\cot A=\dfrac{1}{\tan A}\]
 We are asked in the question to prove the following
\[\tan {{20}^{\circ }}\tan {{80}^{\circ }}\tan {{50}^{\circ }}=\sqrt{3}\]
We see in the left hand side of the proof statement that the two angles ${{20}^{\circ }},{{80}^{\circ }}$ have equal difference of ${{30}^{\circ }}$ from${{50}^{\circ }}$. So if we can express ${{20}^{\circ }},{{80}^{\circ }}$as angle difference from ${{50}^{\circ }}$ or angle sum with ${{50}^{\circ }}$ and then we shall use angle difference formula .So we have;
\[\begin{align}
  & \Rightarrow \tan {{20}^{\circ }}\tan {{80}^{\circ }}\cot {{50}^{\circ }} \\
 & \Rightarrow \tan \left( {{50}^{\circ }}-{{30}^{\circ }} \right)\tan \left( 50+{{30}^{\circ }} \right)\cot {{50}^{\circ }} \\
\end{align}\]
We use the reciprocal relation between tangent and co-tangent $\cot A=\dfrac{1}{\tan A}$ for $A={{50}^{\circ }}$ in the above step to have;
\[\Rightarrow \tan \left( {{50}^{\circ }}-{{30}^{\circ }} \right)\tan \left( 50+{{30}^{\circ }} \right)\dfrac{1}{\tan {{50}^{\circ }}}\]
We use the tangent angle difference formula for $A={{50}^{\circ }},B={{30}^{\circ }}$and have ;
\[\Rightarrow \dfrac{\tan {{50}^{\circ }}-\tan {{30}^{\circ }}}{1+\tan {{50}^{\circ }}\tan {{30}^{\circ }}}\tan \left( 50+{{30}^{\circ }} \right)\dfrac{1}{\tan {{50}^{\circ }}}\]
We use the tangent angle sum formula for $A={{50}^{\circ }},B={{30}^{\circ }}$and have ;
\[\Rightarrow \dfrac{\tan {{50}^{\circ }}-\tan {{30}^{\circ }}}{1+\tan {{50}^{\circ }}\tan {{30}^{\circ }}}\cdot \dfrac{\tan {{50}^{\circ }}+\tan {{30}^{\circ }}}{1-\tan {{50}^{\circ }}\tan {{30}^{\circ }}}\dfrac{1}{\tan {{50}^{\circ }}}\]
We put the value of $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ in the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{\tan {{50}^{\circ }}-\dfrac{1}{\sqrt{3}}}{1+\tan {{50}^{\circ }}\dfrac{1}{\sqrt{3}}}\cdot \dfrac{\tan {{50}^{\circ }}+\dfrac{1}{\sqrt{3}}}{1-\tan {{50}^{\circ }}\dfrac{1}{\sqrt{3}}}\dfrac{1}{\tan {{50}^{\circ }}} \\
 & \Rightarrow \dfrac{\left( \tan {{50}^{\circ }}-\dfrac{1}{\sqrt{3}} \right)\left( \tan {{50}^{\circ }}+\dfrac{1}{\sqrt{3}} \right)}{\left( 1+\dfrac{\tan {{50}^{\circ }}}{\sqrt{3}} \right)\left( 1-\dfrac{\tan {{50}^{\circ }}}{\sqrt{3}} \right)}\dfrac{1}{\tan {{50}^{\circ }}} \\
\end{align}\]
We use algebraic identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ for $a=\tan {{50}^{\circ }},b=\dfrac{1}{\sqrt{3}}$ in the numerator and for $a=1,b=\dfrac{\tan {{50}^{\circ }}}{\sqrt{3}}$ in the denominator of the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( \tan {{50}^{\circ }} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}{{{1}^{2}}-{{\left( \dfrac{\tan {{50}^{\circ }}}{\sqrt{3}} \right)}^{2}}}\dfrac{1}{\tan {{50}^{\circ }}} \\
 & \Rightarrow \dfrac{{{\tan }^{2}}{{50}^{\circ }}-\dfrac{1}{3}}{1-\dfrac{{{\tan }^{2}}{{50}^{\circ }}}{3}}\dfrac{1}{\tan {{50}^{\circ }}} \\
 & \Rightarrow \dfrac{\dfrac{\left( 3{{\tan }^{2}}{{50}^{\circ }}-1 \right)}{3}}{\dfrac{\left( 3-{{\tan }^{2}}{{50}^{\circ }} \right)}{3}\tan {{50}^{\circ }}} \\
 & \Rightarrow \dfrac{3{{\tan }^{2}}{{50}^{\circ }}-1}{3\tan {{50}^{\circ }}-{{\tan }^{3}}{{50}^{\circ }}} \\
\end{align}\]
We see that the above expression is in the form $\tan 3A$ for $A={{50}^{\circ }}$ but in reciprocal form. We take the numerator to the denominator and have;
\[\Rightarrow \dfrac{1}{\dfrac{3\tan {{50}^{\circ }}-{{\tan }^{3}}{{50}^{\circ }}}{3{{\tan }^{2}}{{50}^{\circ }}-1}}\]
We use the tangent triple angle formula for $A={{50}^{\circ }}$ in the denominator to have;
\[\begin{align}
  & \Rightarrow \dfrac{1}{\tan \left( 3\times {{50}^{\circ }} \right)} \\
 & \Rightarrow \dfrac{1}{\tan {{150}^{\circ }}} \\
\end{align}\]
We use the reduction formula about $\pi $ for that is $\tan \left( {{180}^{\circ }}-A \right)=\tan A$ for $A={{30}^{\circ }}$ to have;
\[\begin{align}
  & \Rightarrow \dfrac{1}{\tan \left( {{180}^{\circ }}-{{30}^{\circ }} \right)} \\
 & \Rightarrow \dfrac{1}{\tan {{30}^{\circ }}}=\dfrac{1}{\dfrac{1}{\sqrt{3}}}=\sqrt{3} \\
\end{align}\]
The above result is the same as the result in the right hand side of the given statement. Hence the statement is proved. \[\]

Note: We can alternatively solve by first using reduction formula about $\dfrac{\pi }{2}$ that is $\cot \left( {{90}^{\circ }}-\theta \right)=\tan \theta $ for $\theta ={{40}^{\circ }}$ and then using the tangent triple angle formula $\tan \theta \tan \left( \dfrac{\pi }{3}-\theta \right)\tan \left( \dfrac{\pi }{3}+\theta \right)=\tan 3\theta $ for $\theta ={{20}^{\circ }}$. We must be careful of the common mistake of using $\tan 3A$ formula directly in the step where we used $\tan 3A$ formula in the denominator.