Finding sum of $\sum^{17}_{x=1}\frac{1}{1+\cot^{4}(10^\circ x)}$

Question

Finding sum of $\displaystyle \sum^{17}_{x=1}\frac{1}{1+\cot^{4}(10^\circ x)}$

what i try

assume $\displaystyle A=\sum^{17}_{x=1}\frac{1}{1+\cot^{4}(10^\circ x)}=\sum^{17}_{x=1}\frac{\tan^4(10 ^\circ x)}{1+\tan^4(10^\circ x)}$

$\displaystyle A=\frac{\tan^4(10^\circ)}{1+\tan^4(10^\circ)}+\frac{\tan^4(20^\circ)}{1+\tan^4(20^\circ)}+\cdots \cdots +\frac{\tan^4(170^\circ)}{1+\tan^4(170^\circ)}$

$\displaystyle A=2\frac{\tan^4(10^\circ)}{1+\tan^4(10^\circ)}+2\frac{\tan^4(20^\circ)}{1+\tan^4(20^\circ)}+\cdots +2\frac{1+\tan^4(80^\circ)}{1+\tan^4(80^\circ)}+\frac{\tan^4(90^\circ)}{1+\tan^4(90^\circ)}$

help me to solve please .thanks

Answer 1

Hint $$\displaystyle \sum^{17}_{x=1}\frac{1}{1+\cot^{4}(10^\circ x)}=\sum^{17}_{x=1}\frac{\sin^4(10^\circ x)}{\sin^4(10^\circ x)+\cos^{4}(10^\circ x)}$$

Hint 2 $$\sum^{8}_{x=1}\frac{\sin^4(10^\circ x)}{\sin^4(10^\circ x)+\cos^{4}(10^\circ x)}= \sum^{8}_{x=1}\frac{\cos^4(90^\circ -10^\circ x)}{\cos^4(90^\circ-10^\circ x)+\sin^{4}(90^\circ -10^\circ x)}$$

Hint 3

What is $$\sum^{8}_{x=1}\frac{\sin^4(10^\circ x)}{\sin^4(10^\circ x)+\cos^{4}(10^\circ x)}+\sum^{8}_{x=1}\frac{\cos^4(10^\circ x)}{\sin^4(10^\circ x)+\cos^{4}(10^\circ x)} $$

Hint 4: Use the same trick for $$\sum^{17}_{x=10}\frac{\sin^4(10^\circ x)}{\sin^4(10^\circ x)+\cos^{4}(10^\circ x)}$$