Question

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

• A. $16$
• B. $20$
• C. $12$
• D. $9$

Answer 1

The correct option is C $12$

${\log }_{10}\left(\sin x\right)+{\log }_{10}\left(\cos x\right)=-1\sin x.\cos x=\frac{1}{10}\cdots \left(1\right)$
and ${\log }_{10}(\sin x+\cos x)=\frac{1}{2}({\log }_{10}n-1)\Rightarrow \sin x+\cos x={\left(\frac{n}{10}\right)}^{\frac{1}{2}}\Rightarrow {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=\frac{n}{10}\text{(squaring)}\Rightarrow 1+2\left(\frac{1}{10}\right)=\frac{n}{10}\text{(using equation (1))}\Rightarrow \frac{n}{10}=\frac{12}{10}\Rightarrow n=12$