Question

If ${\log }_{10}\sin x+{\log }_{10}\cos x=-1$ and ${\log }_{10}(\sin x+\cos x)=\frac{{\log }_{10}n-1}{2}$, then the value of $\frac{n}{3}$ is

Answer 1

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

Also, ${\log }_{10}(\sin x+\cos x)=\frac{{\log }_{10}\left(\frac{n}{10}\right)}{2}$

$\Rightarrow {\log }_{10}(\sin x+\cos x{)}^2={\log }_{10}\left(\frac{n}{10}\right)$
$\Rightarrow 1+\sin 2x=\frac{n}{10}$
$\Rightarrow \frac{n}{10}=1+\frac{1}{5}\left[\text{From}\right(1\left)\right]$
$\Rightarrow n=12$
$\therefore \frac{n}{3}=4$