Catch all the cots!

Call the sum $S$ . Use $\displaystyle\sum_{r=a}^bf(r)= \displaystyle\sum_{r=a}^bf(a+b-r)$ ,

$\begin{aligned} S= \sum_{\theta =1^{0}}^{89^{0}}\dfrac{1}{1+cot^{2016}\theta }=& \sum_{\theta =1^{0}}^{89^{0}}\dfrac{1}{1+cot^{2016}(90-\theta) }\\=&\sum_{\theta =1^{0}}^{89^{0}}\dfrac{1}{1+\tan^{2016}\theta }\\=& \sum_{\theta =1^{0}}^{89^{0}}\dfrac{\cot^{2016}\theta}{1+cot^{2016}\theta } \end{aligned}$

Adding we get :

$2S= \sum_{\theta =1^{0}}^{89^{0}}1\implies S=89/2=44.5$

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Alternatively,

$S= \sum_{\theta =1^{0}}^{44^{0}}\frac{1}{1+cot^{2016}\theta }+\underbrace{\dfrac1{1+\cot^{2016} 45^{\circ}}}_{1/2}+\underbrace{ \sum_{\theta =46^{0}}^{89^{0}}\dfrac{1}{1+\cot^{2016}\theta} }_{\text{Use } \cot \theta=\frac1{\cot (90-\theta)}}$

$S= 0.5+\sum_{\theta =1^{0}}^{44^{0}}\frac{1}{1+cot^{2016}\theta }+ \sum_{\theta =1^{0}}^{44^{0}}\frac{cot^{2016}\theta}{1+cot^{2016}\theta }$

$=0.5+ \sum_{\theta =1^{0}}^{44^{0}}(1)$

$= 44.5$