Question

Prove the following trigonometric equation:
\[\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}=3\]

Answer 1

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.