If in a triangle ABC sines of angles A and B satisfy the equation $4{\mathrm{x}}^2-2\sqrt{6}\mathrm{x}+1=0,\text{ then }\cos (\mathrm{A}-\mathrm{B})=$

We have $\sin \mathrm{A}+\sin \mathrm{B}=\sqrt{6}/2\text{ and }\sin \mathrm{Asin}\mathrm{B}=1/4$

$$\begin{gathered} \text{ Let }\mathrm{A}>\mathrm{B}[\because \mathrm{A}\ne \mathrm{B}]. \\ {(\sin A+\sin B)}^2={\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}+2\mathrm{sinAsinB} \\ \Rightarrow {\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}+2\times 1/4=6/4\text{. } \\ \Rightarrow {\sin }^2\mathrm{A}=1-{\sin }^2\mathrm{B} \end{gathered}$$
$\sin \mathrm{A}=\cos \mathrm{B}\Rightarrow \mathrm{B}={90}^{\circ }-\mathrm{A}\Rightarrow \mathrm{A}+\mathrm{B}=\mathrm{C}={90}^{\circ }$
$\text{ Also }\sin \mathrm{Asin}\mathrm{B}=\cos \mathrm{Bsin}\mathrm{B}=1/4\Rightarrow \sin 2\mathrm{B}=1/2\Rightarrow 2\mathrm{B}={30}^{\circ }\text{ or }{150}^{\circ }$
$\Rightarrow \mathrm{B}={15}^{\circ }\text{ or }{75}^{\circ }\Rightarrow \mathrm{B}={15}^{\circ }\text{ and }\mathrm{A}={75}^{\circ }$

cos(A-B)= 1/2