If for $x \in\left(0, \frac{\pi}{2}\right), \log _{10} \sin x+\log _{10} \cos x=-1$ and $\log _{10}(\sin x+\cos x)=\frac{1}{2}\left(\log _{10} n-1\right), n>0$ then the value of $\mathrm{n}$ is equal to :
Correct Option: , 2
$\mathrm{x} \in\left(0, \frac{\pi}{2}\right)$
$\log _{10} \sin x+\log _{10} \cos x=-1$
$\Rightarrow \quad \log _{10} \sin x \cdot \cos x=-1$
$\Rightarrow \quad \sin x \cdot \cos x=\frac{1}{10}$
$\log _{10}(\sin x+\cos x)=\frac{1}{2}\left(\log _{10} n-1\right)$
$\Rightarrow \quad \sin x+\cos x=10\left(\log _{10} \sqrt{n}-\frac{1}{2}\right)=\sqrt{\frac{n}{10}}$
by squaring
$1+2 \sin x \cdot \cos x=\frac{n}{10}$
$\Rightarrow \quad 1+\frac{1}{5}=\frac{n}{10} \quad \Rightarrow \quad \mathrm{n}=12$
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