In a triangle ABC, if  $\frac{a^2+b^2}{a^2-b^2}\sin (A-B)=1$and the triangle is not right angled, then $\cos (A-B)=$

${\sin }^{2}A+{\sin }^{2}B=\sin (A+B)$

$1-{\cos }^{2}A+{\sin }^{2}B=\sin C$

$\Rightarrow {\cos }^{2}A-{\sin }^{2}B=1-\sin C$

$\Rightarrow \cos (A+B)\cos (A-B)=1-\sin C$

$\Rightarrow \cos (A-B)=\frac{1-\sin C}{-\cos C}=\frac{\sin \frac{c}{2}-\cos \frac{c}{2}}{\sin \frac{c}{2}+\cos \frac{c}{2}}$

$\therefore \cos (A-B)=\frac{\tan \frac{c}{2}-1}{\tan \frac{c}{2}+1}=\tan (\frac{c}{2}-\frac{\pi }{4})$