The value of ${\sum }_{k=1}^{10}\left(\sin \frac{2k\pi }{11}+i\cos \frac{2k\pi }{11}\right)$ is:

Let $S={\sum }_{k=1}^{10}\left(\sin \frac{2k\pi }{11}+i\cos \frac{2k\pi }{11}\right)$ $
$S=i{\sum }_{k=1}^{10}\left(\cos \frac{2k\pi }{11}-i\sin \frac{2k\pi }{11}\right)$$Let$ \alpha = \frac{2\pi}{11}$.Then$ e^{i\alpha}$isthe$ 11\text{-th}$rootofunity.Theexpressioninsideis$ e^{-ik\alpha}$.$$S=i{\sum }_{k=1}^{10}(e^{-i\alpha }{)}^k$$Thesumofall$ 11\text{-th}$rootsofunityis$ 0$:$ \sum_{k=0}^{10} e^{-ik\alpha} = 0 \implies 1 + \sum_{k=1}^{10} e^{-ik\alpha} = 0
$${\sum }_{k=1}^{10}{e}^{-ik\alpha }=-1$ Therefore, $S=i(-1)=-i$.