2021/076) Evaluate $\sum^{89}_{n=0}\frac{1}{1+\tan^3(k^\circ)}$

We have $\sum^{89}_{n=0}\frac{1}{1+\tan^3(k^\circ)}$
 $= \sum^{89}_{n=0}\frac{\cos^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{\cos^3(0^\circ)}{\cos^3(0^\circ)+\sin ^3(0^\circ)} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\cos^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ splitting into 4 parts
$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{\sin^3(45^\circ)}{\sin^3(45^\circ)+\sin ^3(45^\circ)}+ \sum^{89}_{n=46}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ as $\sin\,45^\circ = \cos \, 45^\circ$
$=\frac{1}{1+0} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} + \frac{1}{2}+ \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ putting values and using  $\sin\,x^\circ = \cos \,(90^\circ-x)$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{89}_{n=46}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{46}_{n=89}\frac{\cos^3(90^\circ- k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$ reversing order of sum of cos terms
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)} +  \sum^{44}_{n=1}\frac{\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1}\frac{\sin^3(k^\circ)+\cos^3( k^\circ)}{\cos^3(k^\circ)+\sin ^3(k^\circ)}$
$=\frac{3}{2} + \sum^{44}_{n=1} 1= 1.5 + 44 = 45.5 $