Evaluating S would have been much faster if it were $\frac{(m-1)\pi}{4}$
For $0<\theta<\frac{\pi}{2}$.
$$ S=\sum_{m=1}^{11}\frac{1}{\sin\left(\theta+\frac{(m+1)\pi}{4}\right)\sin\left(\theta+\frac{m\pi}{4}\right)}=4\sqrt2$$
A key observation/trick to solving such kinds of problems include substituting the intimidating notations to simpler ones.
Put $\theta+\frac{(m+1)\pi}{4}=t$
$$ S=\sum_{m=1}^{11}\frac{1}{\sin\left(t\right)\sin\left(t-\frac{\pi}{4}\right)}$$
Multiplying and dividing with $sin\frac{\pi}{4}=\frac{1}{\sqrt2}$
$$ S=\sqrt2\sum_{m=1}^{11}\frac{sin\frac{\pi}{4}}{\sin\left(t\right)\sin\left(t-\frac{\pi}{4}\right)}$$
Now, $\frac{\pi}{4}=t+\frac{\pi}{4}-t=t-\left(t-\frac{\pi}{4}\right)$....(1)
Substituting (1) above,
$$ S=\sqrt2\sum_{m=1}^{11}\frac{sin\left(t-\left(t-\frac{\pi}{4}\right)\right)}{\sin\left(t\right)\sin\left(t-\frac{\pi}{4}\right)}$$
using $sin(a-b)=sin(a)cos(b)-cos(a)sin(b)$
$$ S=\sqrt2\sum_{m=1}^{11}\frac{sin(t)cos\left(\left(t-\frac{\pi}{4}\right)\right)-cos(t)sin\left(\left(t-\frac{\pi}{4}\right)\right)}{\sin\left(t\right)\sin\left(t-\frac{\pi}{4}\right)}$$
$$ S=\sqrt2\sum_{m=1}^{11} cot\left({t-\frac{\pi}{4}}\right)-cot(t)$$
Putting back $\theta+\frac{(m+1)\pi}{4}=t$
$$ S=\sqrt2\sum_{m=1}^{11} cot\left({\theta+\frac{m\pi}{4}}\right)-cot\left(\theta+\frac{(m+1)\pi}{4}\right)$$
On telescoping we get,
$$S=\sqrt 2 \left[cot(\theta+\frac{\pi}{4})-cot(\theta+3\pi)\right]=4\sqrt2$$
$$\left[cot(\theta+\frac{\pi}{4})-cot(\theta+3\pi)\right]=4$$
I think you can solve from here for $\theta$
