Mathematics - Basic maths Question with Solution | TestHub

Given $x\in \left(0,\frac{\pi }{2}\right)$

${\log }_{10}\sin x+{\log }_{10}\cos x=-1$

$\Rightarrow {\log }_{10}\sin x.\cos x=-1$

$\Rightarrow \sin x.\cos x=\frac{1}{10}$   $...\left(1\right)$

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

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

By squaring we get, 

${\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=\frac{n}{10}$

$1+2\sin x.\cos x=\frac{n}{10}$

$\Rightarrow 1+\frac{1}{5}=\frac{n}{10}\Rightarrow n=12$