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 'n/3' is

Answer 1

${\log }_{10}\sin x+{\log }_{10}\cos x=-1...\left(a\right)$

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

$Weknowthat{\sin }^{2}x+{\cos }^{2}x=1$

$Using\left(a\right)$

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

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

$Nowapplying{(a+b)}^2=a^2+b^2+2ab$

$Wherea=\sin x,b=\cos x$

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

$n=12$

$Sothevalueof\frac{n}{3}=4$