Question

Find the value of \[\tan \dfrac{\pi }{20}\tan \dfrac{3\pi }{20}\tan \dfrac{5\pi }{20}\tan \dfrac{7\pi }{20}\tan \dfrac{9\pi }{20}\]

Answer 1

Hint: In this question, first identify the pair of complementary angles. Now use the formulas\[\tan \theta =\cot \left( 90-\theta \right)\] and \[\cot \theta =\tan \left( 90-\theta \right)\] and convert the trigonometric ratios. Now, use \[\tan \theta =\dfrac{1}{\cot \theta }\] and cancel the like terms and use \[\tan \dfrac{\pi }{4}=1\] to get the desired result.

Complete step-by-step answer:

In this question, we have to find the value of \[\tan \dfrac{\pi }{20}\tan \dfrac{3\pi }{20}\tan \dfrac{5\pi }{20}\tan \dfrac{7\pi }{20}\tan \dfrac{9\pi }{20}\]. Before proceeding with this question, let us first understand what complementary angles are. Complementary angles are angles whose sum is equal to \[\dfrac{\pi }{2}\]. If we have, \[\angle A+\angle B=\dfrac{\pi }{2}\], then \[\angle A\] and \[\angle B\] are complementary angles of each other.

Similarly, \[\theta \] and \[\left( \dfrac{\pi }{2}-\theta \right)\] are complementary to each other because \[\theta +\dfrac{\pi }{2}-\theta =\dfrac{\pi }{2}\].

So, in trigonometry, we have multiple formulas related to complementary angles and that are the following:

\[\begin{align}

  & \sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta ;\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta \\

 & \cot \left( \dfrac{\pi }{2}-\theta \right)=\tan \theta ;\tan \left( \dfrac{\pi }{2}-\theta \right)=\cot \theta \\

 & \operatorname{cosec}\left( \dfrac{\pi }{2}-\theta \right)=\sec \theta ;\sec \left( \dfrac{\pi }{2}-\theta \right)=\operatorname{cosec}\theta \\

\end{align}\]

Let us now consider the expression given in the question,

\[E=\tan \dfrac{\pi }{20}\tan \dfrac{3\pi }{20}\tan \dfrac{5\pi }{20}\tan \dfrac{7\pi }{20}\tan \dfrac{9\pi }{20}....\left( i \right)\]

In the above expression, we can see that,

\[\dfrac{\pi }{20}+\dfrac{9\pi }{20}=\dfrac{10\pi }{20}=\dfrac{\pi }{2}\]

So, these angles are complementary. By using \[\tan \theta =\cot \left( \dfrac{\pi }{2}-\theta \right)\], we get,

\[\tan \dfrac{\pi }{20}=\cot \left( \dfrac{\pi }{2}-\dfrac{\pi }{20} \right)=\cot \dfrac{9\pi }{20}\]

Similarly, we can see that,

\[\dfrac{3\pi }{20}+\dfrac{7\pi }{20}=\dfrac{10\pi }{20}=\dfrac{\pi }{2}\]

So, these angles are also complementary. So, by again using \[\tan \theta =\cot \left( \dfrac{\pi }{2}-\theta \right)\], we get,

\[\tan \dfrac{3\pi }{20}=\cot \left( \dfrac{\pi }{2}-\dfrac{3\pi }{20} \right)=\cot \dfrac{7\pi }{20}\]

By substituting the value of \[\tan \dfrac{\pi }{20}\] and \[\tan \dfrac{3\pi }{20}\] in the expression (i), we get,

\[E=\cot \dfrac{9\pi }{20}.\cot \dfrac{7\pi }{20}.\tan \dfrac{5\pi }{20}.\tan \dfrac{7\pi }{20}.\tan \dfrac{9\pi }{20}\]

We know that \[\cot \theta =\dfrac{1}{\tan \theta }\]. So, by using this in the above expression, we get,

\[E=\dfrac{1}{\tan \dfrac{9\pi }{20}}.\dfrac{1}{\tan \dfrac{7\pi }{20}}.\tan \dfrac{5\pi }{20}.\tan \dfrac{7\pi }{20}.\tan \dfrac{9\pi }{20}\]

Now, we know that \[\tan \dfrac{5\pi }{20}=\tan \dfrac{\pi }{4}\]. So, by using this and rearranging the terms, we get,

\[E=\dfrac{1}{\tan \dfrac{9\pi }{20}}.\tan \dfrac{9\pi }{20}.\dfrac{1}{\tan \dfrac{7\pi }{20}}.\tan \dfrac{7\pi }{20}.\tan \dfrac{5\pi }{20}.\tan \dfrac{\pi }{4}\]

Now, by canceling the like terms, we get,

\[E=\tan \dfrac{\pi }{4}\]

\[E=\tan \left( \dfrac{\pi }{4}.\dfrac{{{180}^{o}}}{\pi } \right)=\tan {{45}^{o}}\]

	\[\sin \theta \]	\[\cos \theta \]	\[\tan \theta \]	\[\operatorname{cosec}\theta \]	\[\sec \theta \]	\[\cot \theta \]
0	0	1	0	-	1	-
\[\dfrac{\pi }{6}\]	\[\dfrac{1}{2}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{3}}\]	2	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{3}\]
\[\dfrac{\pi }{4}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{\sqrt{2}}\]	1	\[\sqrt{2}\]	\[\sqrt{2}\]	1
\[\dfrac{\pi }{3}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{2}\]	\[\sqrt{3}\]	\[\dfrac{2}{\sqrt{3}}\]	2	\[\dfrac{1}{\sqrt{3}}\]
\[\dfrac{\pi }{2}\]	1	0	-	1	-	0

From the above table, we can see that \[\tan {{45}^{o}}=1\]. So, we get,

E = 1

So, we get the value of

\[\tan \dfrac{\pi }{20}\tan \dfrac{3\pi }{20}\tan \dfrac{5\pi }{20}\tan \dfrac{7\pi }{20}\tan \dfrac{9\pi }{20}=1\]

Note: In these types of questions, it is very important to identify the pair of complementary angles. This question can also be solved in the following way.
\[E=\tan \dfrac{\pi }{20}\tan \dfrac{3\pi }{20}\tan \dfrac{5\pi }{20}\tan \dfrac{7\pi }{20}\tan \dfrac{9\pi }{20}\]
We know that \[\tan \theta =\cot \left( \dfrac{\pi }{2}-\theta \right)\] using this, we get,
\[E=\tan \dfrac{\pi }{20}\tan \dfrac{3\pi }{20}\tan \dfrac{\pi }{4}\cot \left( \dfrac{\pi }{2}-\dfrac{7\pi }{20} \right)\cot \left( \dfrac{\pi }{2}-\dfrac{9\pi }{20} \right)\]

\[E=\tan \dfrac{\pi }{20}\tan \dfrac{3\pi }{20}\tan \dfrac{\pi }{4}\cot \left( \dfrac{\pi }{20} \right)\cot \left( \dfrac{\pi }{20} \right)\]

We know that \[\tan \theta =\dfrac{1}{\cot \theta }\] and \[\tan \dfrac{\pi }{4}=1\]. Using these, we get,

\[E=\dfrac{1}{\cot \dfrac{\pi }{20}}.\cot \dfrac{\pi }{20}.\dfrac{1}{\cot \dfrac{3\pi }{20}}.\cot \dfrac{3\pi }{20}.\tan \dfrac{\pi }{4}\]
\[E=1\]