How to find $B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$ from $\sin\omega+\cos\omega=2\sin\phi$ when $\phi$ is not quadrantal? The problem is as follows:

First let:
$\sin\omega+\cos\omega=2\sin\phi$ and $\phi$ is a not quadrantal
angle.
Then using this relationship find:
$B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$

The alternatives on my book are as follows:
$\begin{array}{ll}
1.&\textrm{0.25}\\
2.&\textrm{0.5}\\
3.&\textrm{1}\\
4.&\textrm{2}\\
\end{array}$
Well what I did to attempt to solve this problem from my precalculus workbook was the following:
I did noticed that on the first expression I could do the following,
$\sin\omega+\cos\omega=2\sin\phi$
$\frac{1}{\sqrt{2}}\sin\omega+\frac{1}{\sqrt{2}}\cos\omega=\frac{2}{\sqrt{2}}\sin\phi$
Then using the sum of angles formula:
$\sin(45^\circ+\omega)=\frac{2}{\sqrt{2}}\sin\phi$
But that's where In got stuck. Now what?
I can't spot a way to make the inner part of the function to be equated with both sides of the equation. In other words, the earlier expression's argument cannot be equated as it is because both sides of the equation are different.
Then there is the other part of the equation which I don't understand. What does it mean not a quadrantal angle?. I mean I do know what a quadrantal is one that is in the standard position and has a measure that is a multiple of $90\circ$. So in short it cannot be $90^\circ$, $180^\circ$, $270^\circ$, $360^\circ$ ... and so on.
But I don't know how to include this information in the solution.
The expression from below doesn't seem to be reduced:
$B=\csc^2 \phi-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$
I could only found that I could:
$B=\frac{1}{\sin^2\phi}-\frac{\cos^2 (45^\circ+\omega)}{\sin^2\phi}$
$B=\frac{1-\cos^2 (45^\circ+\omega)}{\sin^2\phi}$
Can someone help me here with the right approach?. Was my initial intuition right?. I would like someone could help me with that. Please try to include the most step by step as possible so I can catch properly the idea of the necessary manipulation to solve this problem without much fuss.
Can someone help me here please?. I'd really commend someone could guide me in the right path.

First, we have:
$$\sqrt{2}\sin(\phi) = \frac{\sqrt{2}}{2}\sin(\omega) + \frac{\sqrt{2}}{2}\cos(\omega) = \cos\bigg(\frac{\pi}{4}\bigg)\sin(\omega) + \sin\bigg(\frac{\pi}{4}\bigg)\cos(\omega) = \sin\bigg(\frac{\pi}{4}+\omega\bigg)$$
Then:
$$B = \csc^{2}(\phi) -\frac{\cos^{2}\big(\frac{\pi}{4} + \omega\big)}{\sin^{2}(\phi)} =\frac{1}{\sin^{2}(\phi)}-\frac{1 - \sin^{2}\big(\frac{\pi}{4} + \omega\big)}{\sin^{2}(\phi)}=\frac{(\sqrt{2}\sin(\phi))^{2}}{\sin^{2}(\phi)}=\boxed{2}$$
