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.