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

The correct option is B $1$

Consider the following identity
$\tan \alpha \left(\tan \right({90}^0-\alpha )$
$=\tan \alpha \left(\cot \right(\alpha \left)\right)$
$=1$
Therefore the product of $tan$ of complimentary angles is 1 ...(i)
Considering the above given equation
$\tan 1^0\tan 1^0...\tan {89}^0$
$=\tan 1^0\tan {89}^0\tan 2^0\tan {88}^0...\tan {44}^0\tan {46}^0\tan {45}^0$
$=\left(1\right)\left(1\right)...\left(1\right)\left(1\right)$ ...from (i)
$=1$
Hence answer is B