$If for x\in \left(0,\frac{\pi }{2}\right), {\log }_{10}\sin x+{\log }_{10}\cos x=-1 \text{ and }{\log }_{10}(\sin x+\cos x)=\frac{1}{2}\left({\log }_{10}n-1\right),n>0,\text{ then }$$the value of n is equal to$

$\begin{array}{c}{\log }_{10}\sin x+{\log }_{10}\cos x=-1\\ \Rightarrow {\log }_{10}\sin x\cos x=-1\\ \Rightarrow \sin x\cos x=\frac{1}{10}\\ \Rightarrow \sin 2x=\frac{1}{5}\end{array}$

And 

$\begin{array}{c}{\log }_{10}\left(\sin x+\cos x\right)=\frac{1}{2}\left(\log n_{10}-1\right)\\ \Rightarrow {\left(\sin x+\cos x\right)}^2=\frac{n}{10}\\ \Rightarrow 1+\sin 2x=\frac{n}{10}\\ \Rightarrow 1+\frac{1}{5}=\frac{n}{10}\\ \Rightarrow n=10\left(\frac{6}{5}\right)\\ =12\end{array}$