If  $\text{sin}\left(\text{sinx+cosx}\right)\text{=cos}\left(\text{cosx-sinx}\right)\text{,}$  then the largest possible value of   $\frac{\text{π}}{\text{sin\hspace{0.17em}x}}$ is

 $\cos (\cos x-\sin x)=\cos (\frac{\pi }{2}-(\sin x+\cos x))$
 $\therefore \cos x-\sin x=2n\pi \pm (\frac{\pi }{2}-(\sin x+\cos x))$
Taking + ve sign 
  $\text{cosx}-\text{sinx=2nπ+}\frac{\text{π}}{\text{2}}-\text{sinx}-\text{cosx}$
  $\Rightarrow \text{2cosx=2nπ+}\frac{\text{π}}{\text{2}}\Rightarrow \text{cosx = nπ+}\frac{\text{π}}{\text{4}}$
n= 0,  $\cos x=\frac{\pi }{4}\left(\text{only possible values}\right)$
$\sin x=\sqrt{1-\frac{{\pi }^2}{16}}=\frac{\sqrt{16-{\pi }^2}}{4}$ 
 Taking – ve sign 
$\text{cosx}-\text{sinx = 2nπ }-\frac{\text{π}}{\text{2}}-\text{2nπ}\Rightarrow \text{sinx = }\frac{\text{π}}{\text{4}}-\text{nπ}$ 
Only possible  $\sin x=\frac{\pi }{4}$
As   $\frac{\text{π}}{\text{4}}\text{>}\frac{\sqrt{\text{16}}{\text{-π}}^{\text{2}}}{\text{4}}$
 $\Rightarrow \sin x{|}_{\max }=\frac{\pi }{4}$