Question

Evaluate:

$\sum _{r=1}^{89}\log \left(\tan r^o\right)$.

Answer 1

Now,

$\sum _{r=1}^{89}\log \left(\tan r^o\right)$

$=\log \left(\tan 1^o\right)+.....+\log \left(\tan {44}^o\right)+\log \left(\tan {45}^o\right)+\log \left(\tan {46}^o\right)+.....+\log \left(\tan {89}^o\right)$

$=\log \{\tan 1^o.\tan 2^o.....\tan {44}^o.\tan {45}^o.\tan {46}^o....\tan {89}^o\}$

$=\log \{\tan 1^o.\tan 2^o.....\tan {44}^o.1.\cot {44}^o....\cot 1^o\}$ [ Since $\tan \theta =\cot ({90}^o-\theta )$]

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

$=0$.