Question

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 $n$ is equal to :

• A. 20
• B. 12
• C. 9
• D. 16

Answer 1

Answer: (B)

$x\in \left(0,\frac{\pi }{2}\right)$
${\log }_{10}\sin x+{\log }_{10}\cos x=-1$
$\Rightarrow {\log }_{10}\sin x\cdot \cos x=-1$
$\Rightarrow \sin x\cdot \cos x=\frac{1}{10}\dots \dots (1)$
${\log }_{10}(\sin x+\cos x)=\frac{1}{2}\left({\log }_{10}n-1\right)$
$\Rightarrow \sin x+\cos x=10^{\left({\log }_{10}\sqrt{n}-\frac{1}{2}\right)}=\sqrt{\frac{n}{10}}$
by squaring
$1+2\sin x\cos x=\frac{n}{10}$
$\Rightarrow 1+\frac{1}{5}=\frac{n}{10}\Rightarrow n=12$