2016/071) $\tan\, 1 + \tan\, 5 + \tan\, 9 + \cdots + \tan\, 177 =$ (angles in degrees)

we have $\tan\, 45 x  = 1$ has solutions $\tan\, 1 ,  \tan\, 5  \tan\, 9  \cdots  \tan\, 177$ there are 45 solutions
expand tan nx with n = 45 (odd) we get
$\tan\,45 x= \frac{\sum_{n=0}^{22} (-1)^{n}{45 \choose 2n+ 1}\tan^{2n+1}x }{\sum_{n=0}^{22}{45 \choose 2n}(-1)^{n}\tan^{2n}x}$
or $\frac{\sum_{n=0}^{22} (-1)^{n}{45 \choose 2n+ 1}\tan^{2n+1}x }{\sum_{n=0}^{22}{45 \choose 2n}(-1)^{n}\tan^{2n}x}= 1$
or $\sum_{n=0}^{22} (-1)^{n}{45 \choose 2n+ 1}\tan^{2n+1}x -\sum_{n=0}^{22}{45 \choose 2n}(-1)^{n}\tan^{2n}x= 0$
the sum of roots is -ve coefficient of $tan^{44}x$ which is 45
hece  $\tan\, 1 + \tan\, 5 + \tan\, 9 + \cdots + \tan\, 177 =45$