Question

Find the value of
$(\tan 1^o\tan 2^o\tan 3^o........\tan {89}^o)$

Answer 1

As $\tan \theta =\cot (50-\theta )$ and $\tan \theta =\frac{1}{\cot \theta }$

$\Rightarrow \tan 1=\cot 89=\frac{1}{\tan 89}$, $\tan 2=\frac{1}{\tan 88}$$\dots \dots \dots \tan 44=\frac{1}{\tan 46}$

$\Rightarrow$ Expression $=\mathrm{1.1.1...}\dots \tan \mathrm{45......1}=1$.