sequences and series - $cos^2(frac{pi}{101})+cos^2(frac{2pi}{101})+cos^2(frac{3pi}{101})+...+cos^2(frac{100pi}{101})=?$

Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$

My attempt:I've tried it by considering the sum
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$

along with

$$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots +\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$ which gives $100$ as resultant but failed to separate the sum of
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots\\ \dots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$ at last.

I tried the next approach by using de Movire's theorem but failed to separate the real and imaginary part.

I've invested a great amount of time in the so it would be better if someone please come up with an answer.

Answer

$$\cos\left(\frac{k\pi}{101}\right)= \frac{1}{2} \left(e^{i\frac{k\pi}{101}}+e^{-i\frac{k\pi}{101}} \right) \\ \cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right) \\ \sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \sum_{k=1}^{100}\left(e^{2i\frac{k\pi}{101}}+e^{-2i\frac{k\pi}{101}} +2\right)$$

Now,
$$1+\sum_{k=1}^{100}e^{2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{2i\frac{\pi}{101}})^{101}}{1-e^{2i\frac{\pi}{101}}}=0 \\ 1+\sum_{k=1}^{100}e^{-2i\frac{k\pi}{101}}=\sum_{k=0}^{100}\left(e^{-2i\frac{\pi}{101}}\right)^k=\frac{1-(e^{-2i\frac{\pi}{101}})^{101}}{1-e^{-2i\frac{\pi}{101}}}=0$$

Therefore
$$\sum_{k=1}^{100}\cos^2\left(\frac{k\pi}{101}\right)= \frac{1}{4} \left(-1-1+200\right)$$