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

Write each term in exponential form using the identity $$\cos\phi + i\sin\phi = e^{i\phi}$$.

First rewrite the given expression: $$\sin\theta + i\cos\theta = \cos\!\left(\tfrac{\pi}{2}-\theta\right) + i\sin\!\left(\tfrac{\pi}{2}-\theta\right) = e^{\,i\left(\frac{\pi}{2}-\theta\right)}.$$

For $$\theta_k = \dfrac{2k\pi}{11}$$ we therefore have
$$\sin\dfrac{2k\pi}{11} + i\cos\dfrac{2k\pi}{11} = e^{\,i\left(\frac{\pi}{2}-\frac{2k\pi}{11}\right)} = e^{\,i\frac{\pi}{2}}\;e^{-i\frac{2k\pi}{11}} = i\left(e^{-i\frac{2\pi}{11}}\right)^{k}.$$

Thus the required sum becomes
$$S = \sum_{k=1}^{10}\left(\sin\dfrac{2k\pi}{11}+i\cos\dfrac{2k\pi}{11}\right) = i\sum_{k=1}^{10}\left(e^{-i\frac{2\pi}{11}}\right)^{k}.$$

Let $$r = e^{-i\frac{2\pi}{11}}.$$ Then $$r^{11}=e^{-i2\pi}=1,$$ so the eleven 11th roots of unity satisfy $$1 + r + r^{2} + \dots + r^{10}=0.$$

The sum we need is the same series without the first term 1:
$$\sum_{k=1}^{10} r^{k} = \bigl(1 + r + r^{2} + \dots + r^{10}\bigr) - 1 = 0 - 1 = -1.$$

Therefore
$$S = i\bigl(-1\bigr) = -i.$$

Option D which is: -i