`log_(10)sinx+log_(10)cosx=-1 , log_(10)(sinx+cosx)=1/2(log_(10)n-1)` Then n=

Answer

To solve the given problem step by step, we start with the equations provided:

1. **Given Equations**:
\[ \log_{10}(\sin x) + \log_{10}(\cos x) = -1 \]
\[ \log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1) \]

2. **Combine the First Equation**:
Using the logarithmic property that states \(\log a + \log b = \log(ab)\), we can rewrite the first equation:
\[ \log_{10}(\sin x \cdot \cos x) = -1 \]
This implies:
\[ \sin x \cdot \cos x = 10^{-1} = \frac{1}{10} \]

3. **Use the Identity for Sine and Cosine**:
We know from trigonometric identities that:
\[ \sin x \cdot \cos x = \frac{1}{2} \sin(2x) \]
Therefore, we can write:
\[ \frac{1}{2} \sin(2x) = \frac{1}{10} \]
This leads to:
\[ \sin(2x) = \frac{2}{10} = \frac{1}{5} \]

4. **Substituting into the Second Equation**:
Now, we need to evaluate the second equation:
\[ \log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1) \]
We can rewrite the right side:
\[ \frac{1}{2}(\log_{10} n - 1) = \frac{1}{2} \log_{10} n - \frac{1}{2} \]

5. **Express \(\sin x + \cos x\)**:
We can use the identity:
\[ \sin x + \cos x = \sqrt{(\sin^2 x + \cos^2 x) + 2\sin x \cos x} = \sqrt{1 + 2 \cdot \frac{1}{10}} = \sqrt{1 + \frac{1}{5}} = \sqrt{\frac{6}{5}} = \frac{\sqrt{6}}{\sqrt{5}} \]

6. **Substituting Back**:
Now substituting this into the logarithmic equation:
\[ \log_{10}\left(\frac{\sqrt{6}}{\sqrt{5}}\right) = \frac{1}{2} \log_{10} n - \frac{1}{2} \]
This can be rewritten as:
\[ \log_{10}(\sqrt{6}) - \log_{10}(\sqrt{5}) = \frac{1}{2} \log_{10} n - \frac{1}{2} \]

7. **Simplifying the Left Side**:
The left side can be simplified:
\[ \frac{1}{2} \log_{10}(6) - \frac{1}{2} \log_{10}(5) = \frac{1}{2} (\log_{10}(6) - \log_{10}(5)) = \frac{1}{2} \log_{10}\left(\frac{6}{5}\right) \]

8. **Equating Both Sides**:
Now we equate both sides:
\[ \frac{1}{2} \log_{10}\left(\frac{6}{5}\right) = \frac{1}{2} \log_{10} n - \frac{1}{2} \]
Multiplying through by 2 gives:
\[ \log_{10}\left(\frac{6}{5}\right) = \log_{10} n - 1 \]

9. **Solving for \(n\)**:
Rearranging gives:
\[ \log_{10} n = \log_{10}\left(\frac{6}{5}\right) + 1 \]
This can be rewritten as:
\[ \log_{10} n = \log_{10}\left(\frac{6}{5} \cdot 10\right) = \log_{10}\left(\frac{60}{5}\right) = \log_{10}(12) \]
Therefore, we find:
\[ n = 12 \]

Thus, the final answer is:
\[ \boxed{12} \]