Question

Prove the following equation that:
$\tan {{20}^{\circ }}\tan {{30}^{\circ }}\tan {{40}^{\circ }}\tan {{80}^{\circ }}=1$.

Answer 1

Hint: In this given equation, we can firstly use the value of $\tan {{30}^{\circ }}$ that is $\dfrac{1}{\sqrt{3}}$ in place of $\tan {{30}^{\circ }}$. Then we should show $\tan {{40}^{\circ }}$ and $\tan {{80}^{\circ }}$ as $\tan \left( {{60}^{\circ }}-{{20}^{\circ }} \right)$ and $\tan \left( {{60}^{\circ }}+{{20}^{\circ }} \right)$ respectively in order to simplify the Left Hand Side (LHS) of the question by using the formula for tangent of sum of two angles, that is $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Thereafter, we can simplify the Left Hand Side (LHS) of the question by using the formula for tangent of thrice of an angle, that is $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }$. This will lead us to the answer.

Complete step-by-step answer:

In this given question, we are asked to prove that $\tan {{20}^{\circ }}\tan {{30}^{\circ }}\tan {{40}^{\circ }}\tan {{80}^{\circ }}$ is equal to $1$, that is

$\tan {{20}^{\circ }}\tan {{30}^{\circ }}\tan {{40}^{\circ }}\tan {{80}^{\circ }}=1$.

Here, we are going to use the formula for tangent of sum of two angles, that is$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$……………….(1.1) and

the formula for tangent of thrice of an angle, that is $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }$…………………(1.2).

The process of solving is as follows from the Left Hand Side (LHS) to the Right Hand Side (RHS):

$LHS=\tan {{20}^{\circ }}\tan {{30}^{\circ }}\tan {{40}^{\circ }}\tan {{80}^{\circ }}$

$=\dfrac{1}{\sqrt{3}}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{80}^{\circ }}$ (as $\tan {{30}^{\circ }}$=$\dfrac{1}{\sqrt{3}}$)

$=\dfrac{1}{\sqrt{3}}\left[ \tan {{20}^{\circ }}\tan \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\tan \left( {{60}^{\circ }}+{{20}^{\circ }} \right) \right]$


Now, using equation (1.1) we get,
$=\dfrac{1}{\sqrt{3}}\left[ \tan {{20}^{\circ }}\times \dfrac{\tan {{60}^{\circ }}-\tan {{20}^{\circ }}}{1+\tan {{60}^{\circ }}\tan {{20}^{\circ }}}\times \dfrac{\tan {{60}^{\circ }}+\tan {{20}^{\circ }}}{1-\tan {{60}^{\circ }}\tan {{20}^{\circ }}} \right]$

Now, as $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we can multiply the last two terms and obtain

$=\dfrac{1}{\sqrt{3}}\left[ \tan {{20}^{\circ }}\times \dfrac{{{\tan }^{2}}{{60}^{\circ }}-{{\tan }^{2}}{{20}^{\circ }}}{1-{{\tan }^{2}}{{60}^{\circ }}{{\tan }^{2}}{{20}^{\circ }}} \right]$

 The value of $\tan {{60}^{\circ }}=\sqrt{3}$, therefore, using it in the above equation,

$LHS=\dfrac{1}{\sqrt{3}}\left( \dfrac{3\tan {{20}^{\circ }}-{{\tan }^{3}}{{20}^{\circ }}}{1-3{{\tan }^{2}}{{20}^{\circ }}} \right)$

Now, using equation (1.2) with $\theta ={{20}^{\circ }}$, we obtain

$LHS=\dfrac{1}{\sqrt{3}}\times \tan \left( 3\times {{20}^{\circ }} \right)$

$=\dfrac{1}{\sqrt{3}}\times \tan {{60}^{\circ }}$

$=\dfrac{1}{\sqrt{3}}\times \sqrt{3}$

$\begin{align}

  & =1=RHS \\

 & \Rightarrow LHS=RHS \\

\end{align}$

Hence, we have proved the required condition that the Left Hand Side (LHS) is equal to the Right Hand Side (RHS) in the question.

Therefore, we have obtained our proof for the asked equation in the question that is $\tan {{20}^{\circ }}\tan {{30}^{\circ }}\tan {{40}^{\circ }}\tan {{80}^{\circ }}=1$ as true.

Note: We should note that we should keep $\tan {{20}^{\circ }}$ as it is until the end so that we can use equation (1.2) to convert the whole equation in terms of $\tan {{20}^{\circ }}$ whose value is known. Therefore, in these types of questions, we should always try to convert the equations in terms of the angles whose trigonometric identities are known and thus whose value can be put to simplify the equation and hence obtain the answer.