Question

Prove that \[\tan {{1}^{\circ }}\tan {{2}^{\circ }}\tan {{3}^{\circ }}\tan {{4}^{\circ }}\cdots \cdots \tan {{89}^{\circ }}=1\]

Answer 1

Hint: In this question, we are given a product of value of $\tan {{\theta }^{4}}$ where $\theta $ is between 1 to 89. We have to prove it to be equal to 1. For this, we will use the property of $\tan \theta $ that $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $. And then we will apply $\cot \theta =\dfrac{1}{\tan \theta }$. For the first 44 terms, will convert it to $\dfrac{1}{\tan \theta }$ and hence, evaluate our result. We will also use $\tan {{45}^{\circ }}=1$.

Complete step by step answer:
Here we have to prove that, \[\tan {{1}^{\circ }}\tan {{2}^{\circ }}\tan {{3}^{\circ }}\tan {{4}^{\circ }}\cdots \cdots \tan {{89}^{\circ }}=1\]
Here are 89 terms which are difficult to evaluate one by one and then multiply them to obtain our answer, so we will change some of them into $\cot \theta $.
Let us rewrite the left hand side of the equation as \[\tan {{1}^{\circ }}\tan {{2}^{\circ }}\tan {{3}^{\circ }}\cdots \cdots \tan {{44}^{\circ }}\tan {{45}^{\circ }}\tan {{46}^{\circ }}\cdots \cdots \tan {{88}^{\circ }}\tan {{89}^{\circ }}\]
Let us keep the first 44 terms as they are and change terms from \[{{45}^{th}}\] to ${{89}^{th}}$. As we know, $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ so we will use it for the last 44 terms.
As we can see $\tan {{89}^{\circ }}$ can be written as $\tan \left( {{90}^{\circ }}-{{1}^{\circ }} \right)$ therefore, $\tan {{89}^{\circ }}=\cot {{1}^{\circ }}$.
Similarly,
\[\begin{align}
  & \Rightarrow \tan {{88}^{\circ }}=\tan \left( {{90}^{\circ }}-{{2}^{\circ }} \right)=\cot {{2}^{\circ }} \\
 & \Rightarrow \tan {{87}^{\circ }}=\tan \left( {{90}^{\circ }}-{{3}^{\circ }} \right)=\cot {{3}^{\circ }} \\
 & \vdots \\
 & \vdots \\
 & \Rightarrow \tan {{46}^{\circ }}=\tan \left( {{90}^{\circ }}-{{44}^{\circ }} \right)=\cot {{44}^{\circ }} \\
\end{align}\]
Left side of the equation becomes, \[\Rightarrow \tan {{1}^{\circ }}\tan {{2}^{\circ }}\tan {{3}^{\circ }}\cdots \cdots \tan {{44}^{\circ }}\tan {{45}^{\circ }}\cot {{44}^{\circ }}\cdots \cdots \cot {{2}^{\circ }}\cot {{1}^{\circ }}\]
We know that $\cot \theta =\dfrac{1}{\tan \theta }$ therefore,
\[\begin{align}
  & \Rightarrow \cot {{44}^{\circ }}=\dfrac{1}{\tan {{44}^{\circ }}} \\
 & \vdots \\
 & \vdots \\
 & \Rightarrow \cot {{2}^{\circ }}=\dfrac{1}{\tan {{2}^{\circ }}} \\
 & \Rightarrow \cot {{1}^{\circ }}=\dfrac{1}{\tan {{1}^{\circ }}} \\
\end{align}\]
Hence, equation becomes \[\Rightarrow \tan {{1}^{\circ }}\tan {{2}^{\circ }}\tan {{3}^{\circ }}\cdots \cdots \tan {{44}^{\circ }}\tan {{45}^{\circ }}\dfrac{1}{\tan {{44}^{\circ }}}\cdots \cdots \dfrac{1}{\tan {{2}^{\circ }}}\dfrac{1}{\tan {{1}^{\circ }}}\]
Now, $\tan {{1}^{\circ }}\text{ and }\dfrac{1}{\tan {{1}^{\circ }}}$ will cancelled out and similarly other 43 terms will be too. So we are left with $\tan {{45}^{\circ }}$.

We know, $\tan {{45}^{\circ }}=1$ therefore, the left side of the equation becomes equal to 1 which is equal to the right side of the equation. Hence proved.

Note: Students should be careful while splitting the terms and take care that 44 terms are cancelled by 44 terms only. Hence, $\tan {{90}^{\circ }}$ is not multiplied because $\tan {{90}^{\circ }}$ is not defined. Students should remember all trigonometric formulas such as $\tan \left( {{90}^{\circ }}-\theta \right),\tan \left( {{90}^{\circ }}+\theta \right)$ and take care of signs as well.