Question

If for $\mathrm{x}\in \left(0,\frac{\mathrm{\pi }}{2}\right),{\log }_{10}\sin \mathrm{x}+{\log }_{10}\cos \mathrm{x}=-1$ and ${\log }_{10}(\sin \mathrm{x}+\cos \mathrm{x})=\frac{1}{2}({\log }_{10}\mathrm{n}-1),\mathrm{n}>0$, then the value of $n$ is equal to:

• A. 16
• B. 20
• C. 12
• D. 9

Answer 1

The correct option is C

12

Explanation of the given option:

Step 1: Apply the formlula $\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{a}\mathbf{+}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{b}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{(}\mathbf{a}\mathbf{\times }\mathbf{b}\mathbf{)}$

Given equation, ${\log }_{10}\sin \mathrm{x}+{\log }_{10}\cos \mathrm{x}=-1$

$$\begin{gathered} \therefore {\log }_{10}\sin \mathrm{x}+{\log }_{10}\cos \mathrm{x}=-1 \\ \Rightarrow {\log }_{10}(\mathrm{sinx}.\mathrm{cosx})=-1 \\ \Rightarrow \sin \mathrm{x}.\cos \mathrm{x}={10}^{-1} \\ \Rightarrow \sin \mathrm{x}.\cos \mathrm{x}=\frac{1}{10} \end{gathered}$$

Given equation: ${\log }_{10}(\sin \mathrm{x}+\cos \mathrm{x})=\frac{1}{2}({\log }_{10}\mathrm{n}-1),\mathrm{n}>0$

$$\begin{gathered} \therefore {\log }_{10}\left(\sin \mathrm{x}+\cos \mathrm{x}\right)=\frac{1}{2}({\log }_{10}\mathrm{n}-1) \\ =\frac{1}{2}({\log }_{10}\mathrm{n}-{\log }_{10}10)\left[\because {\log }_{10}=1\right] \\ \therefore {\log }_{10}(\sin \mathrm{x}+\cos \mathrm{x})=\frac{1}{2}{\log }_{10}\left(\frac{\mathrm{n}}{10}\right)\left[\because {\log }_{\mathrm{a}}\mathrm{x}-{\log }_{\mathrm{a}}\mathrm{y}={\log }_{\mathrm{a}}\frac{\mathrm{x}}{\mathrm{y}}\right] \\ \Rightarrow \sin \mathrm{x}+\cos \mathrm{x}={10}^{\frac{1}{2}{\log }_{10}\left(\frac{\mathrm{n}}{10}\right)} \end{gathered}$$

Step 2: Squaring on both sides,

$$\begin{gathered} \therefore {\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}+2\sin \left(x\right)\cos \left(x\right)={10}^{{\log }_{10}\frac{\mathrm{n}}{10}} \\ \Rightarrow 1+\frac{2}{10}=\frac{\mathrm{n}}{10}\left[\because {\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}=1\mathbf{and}\mathrm{sinx}.\mathrm{cosx}=\frac{1}{10},\mathbf{and}{\mathrm{a}}^{{\log }_{\mathrm{a}}\mathrm{k}}=\mathrm{k}\right] \\ \Rightarrow \frac{10+2}{10}=\frac{\mathrm{n}}{10} \\ \Rightarrow \frac{12}{10}=\frac{\mathrm{n}}{10} \\ \Rightarrow \mathrm{n}=12 \end{gathered}$$

Hence, the correct answer is an option (C).