Question

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

• A. $0.8$
• B. $1.2$
• C. $4$
• D. None of these

Answer 1

The correct option is D None of these

$\begin{array}{c}Lo{g}_{10}\sin x+{\log }_{10}\cos x=-1\\ \Rightarrow {\log }_{10}\left(\sin x+\cos x\right)=-1\\ \Rightarrow {\log }_{10}\left(\sin x+\cos x\right)=\frac{\left({\log }_{10}n\right)-1}{2}\\ \Rightarrow -2={\log }_{10}n-1\\ \Rightarrow -1={\log }_{10}n\\ \Rightarrow n={10}^{-1}\\ n=0.1\end{array}$