Express $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n $$ in the form $ x + iy$
$$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n = \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \cdot \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta} \right)^n $$
$$ \left( \frac{\left( 1 + \sin\theta + i\cos\theta \right)^2}{\left(1 + \sin\theta\right)^2 - \left(i\cos\theta\right)^2} \right)^n = \left(\frac{\sin^2\theta}{1 + \sin\theta}+i\cos\theta\right)^n$$
then, if necessary, we can further expand via Newton's Binomial Theorem:
$$ \left( x + y \right)^n = {{n}\choose{k}}x^{n-k}y^k$$
Am I missing something?
Go through exponential form.
$$\sin \theta+ i \cos\theta = i e^{-i\theta}$$ $$\sin \theta- i \cos\theta = -i e^{+i\theta}$$
Your expression is: $$\left(\frac{1+ie^{-i\theta}}{1-ie^{i\theta}}\right)^n=$$ $$\left(\frac{(i)e^{-i\theta}(-i)e^{i\theta}(1+ie^{-i\theta})}{1-ie^{i\theta}}\right)^n=$$ $$\left(\frac{i e^{-i\theta}(-ie^{i\theta}+1)}{1-ie^{i\theta}}\right)^n=$$ $$\left(i e^{-i\theta}\right)^n=\left( e^{-i\theta+i\pi/2}\right)^n=e^{in(\pi/2-\theta)}=\cos (n(\pi/2-\theta))+i\sin(n(\pi/2-\theta))$$
You'll notice the most awkward part is that sine and cosine are exchanged compared to the Euler's formula. For that reason, using a new variable $u=\frac{\pi}{2}-\theta$ for most of the calculation makes things much cleaner and show the true beauty of this solution:
$$\sin\theta+i\cos\theta=\cos u +i \sin u = e^{iu}$$ $$\sin\theta-i\cos\theta=\cos u -i \sin u = e^{-iu}$$ Now you don't have the awkward $i$'s there: $$\left(\frac{1+e^{iu}}{1+e^{-iu}}\right)^n=\left(e^{iu}\right)^n=e^{inu}=\cos nu+i\sin nu$$