Question

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

Answer 1

${\log }_{10}\sin x+{\log }_{10}\cos x=-1$
$\Rightarrow {\log }_{10}\left(\sin x\cos x\right)=-1$
$\Rightarrow {\log }_{10}\left(\frac{\sin 2x}{2}\right)=-1$
$\Rightarrow \frac{\sin 2x}{2}=\frac{1}{10}$
$\Rightarrow \sin 2x=\frac{1}{5}\cdots \left(1\right)$

And, ${\log }_{10}(\sin x+\cos x)=\frac{\left({\log }_{10}n\right)-1}{2}$
$\Rightarrow 2{\log }_{10}(\sin x+\cos x)={\log }_{10}n-{\log }_{10}10$
$\Rightarrow {\log }_{10}(\sin x+\cos x{)}^2={\log }_{10}\left(\frac{n}{10}\right)$
$\Rightarrow (\sin x+\cos x{)}^2=\frac{n}{10}$
$\Rightarrow 1+\sin 2x=\frac{n}{10}$
Using equation $\left(1\right)$, we get
$1+\frac{1}{5}=\frac{n}{10}$
$\Rightarrow n=12$