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# Prove the following trigonometric equation:$\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}=3$  Verified
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Hint: First of all, as we know that we can substitute ${{40}^{o}}=\left( {{60}^{o}}-{{20}^{o}} \right)$ and ${{80}^{o}}=\left( {{60}^{o}}+{{20}^{o}} \right)$. Then use the formula of $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ and $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$.

The expression in the question to be proved is given as
$\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}=3$
Let us consider the LHS of the given expression as below,
$A=\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}$
Since, we know that,
$\tan {{60}^{o}}=\sqrt{3}$
Therefore, by putting the value of $\tan {{60}^{o}}$ in the above expression, we get,
$A=\tan {{20}^{o}}.\tan {{40}^{o}}.\sqrt{3}.\tan {{80}^{o}}$
We can also write the above expression as
$A=\sqrt{3}.\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{80}^{o}}$
Now, we can know that ${{40}^{o}}=\left( {{60}^{o}}-{{20}^{o}} \right)$ and ${{80}^{o}}=\left( {{60}^{o}}+{{20}^{o}} \right)$.
So, we get the above expression as,
$A=\sqrt{3}.\tan {{20}^{o}}.\tan \left( {{60}^{o}}-{{20}^{o}} \right).\tan \left( {{60}^{o}}+{{20}^{o}} \right)$
Since, we know that
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}$
And,
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$
Therefore, by applying the above formulas, we get the above expression as
$A=\left( \sqrt{3} \right).\left( \tan {{20}^{o}} \right)\left[ \dfrac{\tan {{60}^{o}}-\tan {{20}^{o}}}{1+\tan {{60}^{o}}.\tan {{20}^{o}}} \right].\left[ \dfrac{\tan {{60}^{o}}+\tan {{20}^{o}}}{1-\tan {{60}^{o}}\tan {{20}^{o}}} \right]$
Since, we know that,
$\tan {{60}^{o}}=\sqrt{3}$
Therefore, by putting the value of $\tan {{60}^{o}}$ in the above expression, we get
$A=\left( \sqrt{3} \right).\left( \tan {{20}^{o}} \right)\left[ \dfrac{\sqrt{3}-\tan {{20}^{o}}}{1+\sqrt{3}.\tan {{20}^{o}}} \right].\left[ \dfrac{\sqrt{3}+\tan {{20}^{o}}}{1-\sqrt{3}\tan {{20}^{o}}} \right]$
We can also write the above expression as,
$A=\dfrac{\sqrt{3}\tan {{20}^{o}}.\left( \sqrt{3}-\tan {{20}^{o}} \right)\left( \sqrt{3}+\tan {{20}^{o}} \right)}{\left( 1-\sqrt{3}.\tan {{20}^{o}} \right)\left( 1+\sqrt{3}\tan {{20}^{o}} \right)}$
Since we know that
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Therefore, by applying this formula in the above expression, we get,
$A=\sqrt{3}\left( \tan {{20}^{o}} \right).\left[ \dfrac{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \tan {{20}^{o}} \right)}^{2}}}{{{\left( 1 \right)}^{2}}-{{\left( \sqrt{3}\tan {{20}^{o}} \right)}^{2}}} \right]$
By simplifying the above equation, we get,
$\Rightarrow A=\sqrt{3}\left( \tan {{20}^{o}} \right).\left[ \dfrac{3-{{\tan }^{2}}{{20}^{o}}}{1-3{{\tan }^{2}}{{20}^{o}}} \right]$
Now, we will take $\tan {{20}^{o}}$ inside the bracket
We get,
$\Rightarrow A=\sqrt{3}\left[ \dfrac{3\tan {{20}^{o}}-{{\tan }^{3}}{{20}^{o}}}{1-3{{\tan }^{2}}{{20}^{o}}} \right]$
Since, we know that
$\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }=\tan 3\theta$
Therefore, by applying the above formula, we get,
$A=\sqrt{3}\left[ \tan 3.\left( {{20}^{o}} \right) \right]$
We can also write the above expression as
$A=\sqrt{3}\left[ \tan {{60}^{o}} \right]$
Since, we know that $\tan {{60}^{o}}=\sqrt{3}$, therefore by putting the value of $\tan {{60}^{o}}$ in the above expression we get,
$A=\sqrt{3}.\sqrt{3}$
Therefore, A = 3 = RHS
Hence, we proved that the value of $\tan {{20}^{o}}\tan {{40}^{o}}\tan {{60}^{o}}\tan {{80}^{o}}=3$.

Note: Here by looking at the terms like $\tan {{20}^{o}},\tan {{40}^{o}}$ and $\tan {{60}^{o}}$, students often make this mistake of using formulas of double angles that is $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ which makes the solution lengthy and does not lead to the desired result.