Question

The value of $\tan {1^0}.\tan {2^0}.\tan {3^0}........\tan {89^0}$ is
A. 0
B. 1
C. Infinity
D. None

Answer 1

Hint: Here we will find the number of terms in the given equation and apply trigonometry formulae to find the value.

Complete step-by-step answer:

As you know that $\tan (90 - \theta ) = \cot \theta {\text{ & }}\tan \theta .\cot \theta = 1$
So you have to find out the value of $\tan {1^0}.\tan {2^0}.\tan {3^0}........\tan {89^0}$.
$
   \Rightarrow \tan {1^0}.\tan {2^0}.\tan {3^0}........\tan {87^0}.\tan {88^0}.\tan {89^0} \\
   \Rightarrow \tan {1^0}.\tan {2^0}.\tan {3^0}........\tan ({90^0} - {3^0}).\tan ({90^0} - {2^0}).\tan ({90^0} - {1^0}) \\
   \Rightarrow \tan {1^0}.\tan {2^0}.\tan {3^0}........\cot {3^0}.\cot {2^0}.\cot {1^0} \\
$
The number of terms from (1, 2, 3, 4………………………to 89) will be
First term is 1, last term is 89, and common difference is 1.It forms an A.P
$
  89 = 1 + (n - 1)1 \\
  n = 89 \\
$
Which is odd, Therefore mid-term of series is
$ \Rightarrow \dfrac{{1 + 89}}{2} = 45$
$
   \Rightarrow \tan {1^0}.\tan {2^0}.\tan {3^0}........\cot {3^0}.\cot {2^0}.\cot {1^0} \\
   \Rightarrow (\tan {1^0} \times \cot {1^0})(\tan {2^0} \times \cot {2^0})(\tan {3^0} \times \cot {3^0})...........\tan {45^0} \\
   \Rightarrow 1 \times 1 \times 1.............. \times \tan {45^0} \\
   \Rightarrow \tan {45^0} = 1 \\
 $
So, the correct answer is option B.

Note: In this type of question always remember trigonometry properties, it will help you in finding your desired answers.