Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first part, I tried realising LHS so I got $$LHS=\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2+\cos^2\theta}=\frac{1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta}{2+2\sin\theta}.$$
but now I am stuck :( . Any help would be greatly appreciated, thanks!
 A: I write
$i\cos \theta + \sin \theta = i(\cos \theta -i\sin \theta) = i(\cos (-\theta) + i\sin (-\theta)) = ie^{-i\theta}; \tag 1$
then
$\sin \theta -i\cos \theta = -i(\cos \theta + i\sin \theta) = -ie^{i \theta}; \tag 2$
we have
$(1 + \sin \theta - i\cos \theta)(\sin \theta + i\cos \theta) = (1 - ie^{i \theta})ie^{-i\theta} = ie^{-i\theta} + 1 = 1 + \sin \theta + i\cos \theta, \tag 3$
simply a slightly altered form equivalent to the desired result; some minor re-arrangemens yield
$\dfrac{ 1 + \sin \theta + i\cos \theta}{ 1 + \sin \theta - i\cos \theta } = \dfrac{1 + ie^{-i\theta}}{1 - ie^{i \theta}} = ie^{-i\theta} = \sin \theta + i\cos \theta, \tag 4$
the equation in the form given.  
We also wish to show that
$\left (1+\sin\dfrac{\pi}{5}+i\cos\dfrac{\pi}{5} \right )^5+i \left ( 1+\sin\dfrac{\pi}{5}-i\cos\dfrac{\pi}{5} \right )^5=0; \tag 5$
if we set
$\theta = \dfrac{\pi}{5}, \tag 6$
then it follows that
$\left ( \dfrac{ 1 + \sin \theta + i\cos \theta}{ 1 + \sin \theta - i\cos \theta } \right )^5 = (\sin \theta + i \cos \theta)^5$
$= (ie^{-i\theta})^5 = i^5 e^{-i 5 \theta} = ii^4 e^{-i 5(\pi / 5)} = ie^{-i \pi} = -i; \tag 7$
therefore,
$\dfrac{( 1 + \sin \theta + i\cos \theta)^5}{( 1 + \sin \theta - i\cos \theta )^5} = \left ( \dfrac{ 1 + \sin \theta + i\cos \theta}{ 1 + \sin \theta - i\cos \theta } \right )^5 = (\sin \theta + i \cos \theta)^5 = -i, \tag 8$
whence
$( 1 + \sin \theta + i\cos \theta)^5 + i ( 1 + \sin \theta - i\cos \theta )^5 = 0; \tag 9$
if we substitute (6) into this equation we find that in concrete terms
$\left ( 1 + \sin \dfrac{\pi}{5} + i\cos \dfrac{\pi}{5} \right )^5 + i \left ( 1 + \sin \dfrac{\pi}{5} - i\cos \dfrac{\pi}{5} \right )^5 = 0, \tag{10}$ 
as per request.
A: From numerator of your solution
$$1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta$$
$$(1+2\sin\theta+\sin^2\theta-\cos^2\theta)+(2i\cos\theta+2i\cos\theta\sin\theta)$$
$$(1+2\sin\theta+\sin^2\theta-1+\sin^2\theta)+2i\cos\theta(1+\sin\theta)$$
$$(2\sin\theta)(1+\sin\theta)+2i\cos\theta(1+\sin\theta)$$
$$2(1+\sin\theta)(\sin\theta+i\cos\theta)$$
A: Hint:
$$\frac{1+sin\theta + icos\theta}{1+sin\theta-icos\theta}=sin\theta+icos\theta \iff 1+sin\theta + icos\theta =\underbrace{(1+sin\theta-icos\theta)(sin\theta+icos\theta)}_{R}$$
so calculate the right side and...
$$ R = sin\theta+icos\theta + \underbrace{sin^2\theta-i^2cos ^2\theta}_{=1}$$
A: Replace $\theta$ with $90^\circ-2y$
$$F=\dfrac{1+\cos2y+i\sin2y}{1+\cos2y-i\sin2y}$$
Using Double angle formula,
$$F=\dfrac{2\cos^2y+2i\sin y\cos y}{2\cos^2y-2i\sin y\cos y}$$
If $\cos y\ne0$
$$F=\dfrac{\cos y+i\sin y}{\cos y-i\sin y}=(\cos y+i\sin y)^2=\cos2y+i\sin2y$$
