Explanation:

From the given series,
tan1° tan 2° tan 3° ........ tan 89°
we can write,    tan 89° = tan (90 – 1) = cot 1°          [Bcz tan (90° – Θ) = cot Θ]
Similarly,
tan 88° = cot 2° , tan 87° = cot 3°, .......up to tan 46° = cot 44°
Then middle term is tan 45° = 1
So, the series will be
tan 1° . tan 2° . tan 3°, ....... tan 44° . tan 45°. cot 44°,......... cot1°
The terms wtih tanΘ and cotΘ will be cancelled out by [tan Θ . cot Θ = 1]
So, the remaining term is tan 45° = 1.