If ${\log }_{10}\sin \mathrm{x}+{\log }_{10}\cos \mathrm{x}=-1$ and ${\log }_{10}(\sin \mathrm{x}+\cos \mathrm{x})=\frac{\left({\log }_{10}\mathrm{n}\right)-1}{2}$ then the value of n is

$\mathrm{Given} {\log }_{10}\left(\frac{\sin 2\mathrm{x}}{2}\right)=-1$

$\begin{array}{l}\mathrm{or} \frac{\sin 2\mathrm{x}}{2}=\frac{1}{10}\\ \mathrm{or} \sin 2\mathrm{x}=\frac{1}{5}\\ \mathrm{also} {\log }_{10}(\sin \mathrm{x}+\cos \mathrm{x})=\frac{{\log }_{10}\left(\frac{\mathrm{n}}{10}\right)}{2}\\ \mathrm{or} {\log }_{10}(\sin \mathrm{x}+\cos \mathrm{x}{)}^2={\log }_{10}\left(\frac{\mathrm{n}}{10}\right)\\ \mathrm{or} 1+\sin 2\mathrm{x}=\frac{\mathrm{n}}{10}\\ \mathrm{or} 1+\frac{1}{5}=\frac{\mathrm{n}}{10}\\ \mathrm{or} \frac{6}{5}=\frac{\mathrm{n}}{10}\\ \therefore \mathrm{n}=12\end{array}$