Express a complex number in modulus amplitude form Express a complex number in modulus amplitude form
$\displaystyle 1+\sin \alpha +i\cos \alpha $
My Attempt: 
$\displaystyle r\cos \theta= 1+\sin \alpha $
$\displaystyle r\sin \theta= \cos \alpha $
Squaring and adding.. $\displaystyle r^2= (1+\sin \alpha)^2+ \cos^2 \alpha$
$\displaystyle r^2= 2(1+\sin \alpha) $
$\displaystyle \tan \theta = \frac{\cos \alpha}{1+\sin \alpha} $
How to break up $\displaystyle 1+\sin \alpha $?

It is better to solve it this way:
$$\begin{aligned}
1+\sin\alpha+i\cos \alpha &=1+\cos\left(\frac{\pi}{2}-\alpha\right)+i\sin\left(\frac{\pi}{2}-\alpha\right)\\
& \stackrel{*}{=}2\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+i2\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\\
&=2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\left(\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+i\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\right)\\
&=2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)e^{i\left(\pi/4-\alpha/2\right)}\\
\end{aligned}$$
$(*)$, In this step I used the following formulas: 
$\cos(2x)=2\cos^2x-1$ and $\sin(2x)=2\sin x\cos x$

In your method, I am not seeing how you get that expression for $\tan\theta$. Rather it should be $\tan\theta=\dfrac{\cos\alpha}{1+\sin\alpha}$. Hence,
$$\tan\theta=\frac{\sin\left(\frac{\pi}{2}-\alpha\right)}{1+\cos\left(\frac{\pi}{2}-\alpha\right)}=\frac{2\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$$
$$\Rightarrow \tan\theta=\tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$$
