Question

The value of $\tan 1^\circ .\tan 2^\circ .\tan 3^\circ ...\tan 89^\circ =$
A.0
B) 1
C) -1
D) 2

Answer 1

Hint : There are many ways to solve this question, first of all, we will break above series from $\tan 1^\circ$ to $\tan 44^\circ$ and $\tan 46^\circ$ to $\tan 89^\circ$ and keep $\tan 45^\circ$ separate from both the series as we know that the value of $\tan 45^\circ$ is 1. Then we will convert the second series from $\tan 46^\circ$ into $\tan (90 - 44)^\circ$ to make it a series of $\cot$ , so that we can cancel out $\tan x$ with $\cot x$ .

Complete step-by-step answer :
$\tan 1^\circ .\tan 2^\circ .\tan 3^\circ ...\tan 89^\circ$
We will break above question into two series and keep $\tan 45^\circ$ separate from both series as shown below:
$(\tan 1^\circ .\tan 2^\circ .\tan 3^\circ ...\tan 44^\circ )(\tan 45^\circ )(\tan 46^\circ \tan 47^\circ \tan 48^\circ ...\tan 89^\circ )$
We will keep first series and $\tan 45^\circ$ as it is and solving rest of the series as we can write $\tan 46^\circ = \tan (90 - 44)^\circ$ and so on
$= (\tan 1^\circ .\tan 2^\circ .\tan 3^\circ ...\tan 44^\circ )(\tan 45^\circ )\tan (90^\circ - 44^\circ )\tan (90^\circ - 43^\circ )(90^\circ - 42^\circ )...\tan (90^\circ - 1^\circ )$
We know that we have an identity in trigonometry $\tan \left( {90 - x} \right) = \cot x$ , as $90^\circ - x$ will belongs to first quadrant, where $\tan \left( {90 - x} \right)$ will be $\cot x$ and sign will remain positive, as all trigonometry function are positive in first quadrant. Therefore we can write $\tan (90 - 44)^\circ = \tan 44^\circ$ and so on.
$= (\tan 1^\circ .\tan 2^\circ .\tan 3^\circ ...\tan 44^\circ )(\tan 45^\circ )(\cot 44^\circ \cot 43^\circ \cot 42^\circ ...\cot 1^\circ )$
Now, we will take a similar angle together as shown below, as we can put an identity into it.
$= (\tan 1^\circ .\cot 1^\circ .\tan 2^\circ .\cot 2^\circ .\tan 3^\circ .\cot 3^\circ ...\tan 44^\circ .\cot 44^\circ )(\tan 45^\circ )$
Here, we will put the value of $\tan 45^\circ = 1$ and also apply an identity which we know $\tan x.\cot x = 1$
$= 1.1.1 \ldots .1.1 \Rightarrow {1^n}$
=1
Hence the result of the above series is 1. So option b is the right option.
So, the correct answer is “Option B”.

Note : Second method to solve above question:
We know that we have an identity in trigonometry $\tan \left( {90 - x} \right) = \cot x$
Therefore $\tan 1^\circ = \cot 89^\circ$ and $\tan 2^\circ = \cot 88^\circ$ so on, we will convert above series from $\tan 1^\circ$ to $\tan 44^\circ$ into cot series.
$= (\cot 89^\circ .\cot 88^\circ .\cot 87^\circ ...\cot 46^\circ )(\tan 45^\circ )(\tan 46^\circ \tan 47^\circ \tan 48^\circ ...\tan 89^\circ )$
We change tan into cot as we know that $\tan x = \dfrac{1}{{\cot x}}$ as shown below.
$= (\cot 89^\circ .\cot 88^\circ .\cot 87^\circ ...\cot 46^\circ )(\tan 45^\circ )(\dfrac{1}{{\cot 46^\circ }}.\dfrac{1}{{\cot 47^\circ }}.\dfrac{1}{{\cot 48^\circ }}...\dfrac{1}{{\cot 89^\circ }})$
Now, we will take similar angle together as shown below
$\Rightarrow (\cot 89^\circ .\dfrac{1}{{\cot 89^\circ }}\cot 88^\circ \dfrac{1}{{\cot 88^\circ }}.\cot 87^\circ .\dfrac{1}{{\cot 87^\circ }}...\cot 46^\circ .\dfrac{1}{{\cot 46^\circ }})(\tan 45^\circ )$
Here, we will put the value of $\tan 45^\circ = 1$ and also cancel out similar terms
$= 1.1.1 \ldots .1.1 \Rightarrow {1^n}$
$= 1$
Hence the result of the above series is 1. So option b is the right option.