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)

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