Nice exercise about the sum of log (tangents)

Exercise:

Determine the value of $\sum\limits_{n=1}^{89}{\log \left( \tan \left( n \right) \right)}$

Solution: we know that, $\log \prod\limits_{n=1}^{m}{{{a}_{n}}}=\sum\limits_{n=1}^{m}{\log {{a}_{n}}}$

$\sum\limits_{n=1}^{89}{\log \left( \tan \left( n \right) \right)=\log \left( \tan 1 \right)+\log \left( \tan 2 \right)+....+\log \left( \tan 88 \right)+\log \left( \tan 89 \right)}$

                     $=\log \left( \prod\limits_{n=1}^{89}{\tan \left( n \right)} \right)$

                    $=\log \left( \tan 1\times \tan 2\times \tan 3\times .....\times \tan 88\times \tan 89 \right)$

                   $=\log \left( \tan 1\times \tan 2\times ....\times \tan 44\times \tan 45\times \tan 46\times .....\times \tan 88\times \tan 89 \right)$

But $\tan \theta =\frac{1}{\cot \theta }=\frac{1}{\tan \left( 90-\theta  \right)}$

So observe that,

$\tan 1=\frac{1}{\tan \left( 90-1 \right)}=\frac{1}{\tan 89}$ , $\tan 2=\frac{1}{\tan \left( 90-2 \right)}=\frac{1}{\tan 88}$

$\tan 44=\frac{1}{\tan \left( 90-44 \right)}=\frac{1}{\tan 46}$ , $\tan 45=1$

So \(\sum\limits_{n=1}^{89}{\log \left( \tan \left( n \right) \right)=}\log \left( \frac{1}{\tan 89}\times \frac{1}{\tan 88}\times ....\times \frac{1}{\tan 46}\times 1\times \tan 46\times ....\times \tan 88\times \tan 89 \right)\)

                         $=\log \left( 1 \right)=0$