$\frac{{\sin }^2{135}^{\circ }+{\sin }^2{45}^{\circ }}{\mathrm{tg}{45}^2-\mathrm{ctg}{135}^{\circ }}=$  

 

$=\frac{{\left(\sin {90}^{\circ }+{45}^{\circ }\right)}^2+{\sin }^2{45}^{\circ }}{\mathrm{tg}{45}^2-\mathrm{ctg}\left({90}^{\circ }+{45}^{\circ }\right)}=$  

 

$=\frac{{\cos }^2{45}^{\circ }+{\sin }^2{45}^{\circ }}{\mathrm{tg}{45}^2-\left(-\mathrm{tg}{45}^{\circ }\right)=}$  

 

$=\frac{1}{1-\left(-1\right)}=\frac{1}{2}\in \mathbb{Q}$