Question

If ${\log }_{10}\left(\sin (x+\frac{\pi }{4})\right)=\frac{{\log }_{10}6-1}{2}$, then the value of ${\log }_{10}\left(\sin x\right)+{\log }_{10}\left(\cos x\right)$ will be

• A. $-1$
• B. $-2$
• C. $2$
• D. $1$

Answer 1

The correct option is A $-1$

$2{\log }_{10}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)={\log }_{10}\left(\frac{6}{10}\right)$

$\Rightarrow {\log }_{10}\left(\frac{1+2\sin x\cos x}{2}\right)={\log }_{10}\left(\frac{6}{10}\right)[\because m\log a=\log a^m]$

$\Rightarrow \frac{1}{2}+\sin x\cos x=\frac{3}{5}\Rightarrow \sin x\cos x=\frac{1}{10}$

$\Rightarrow {\log }_{10}\left(\sin x\right)+{\log }_{10}\left(\cos x\right)={\log }_{10}\left(\sin x\cos x\right)=-1[\because \log a+\log b=\log (ab\left)\right]$

Hence, $\left(A\right)$ is the correct answer.