If for x in left(0, frac{pi}{2}right),log_{10 | vedclass

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(B) Given $x \in \left(0, \frac{\pi}{2}\right)$.From the first equation: $\log_{10} \sin x + \log_{10} \cos x = -1$$\Rightarrow \log_{10}(\sin x \cos

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$12$

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(B) Given $x \in \left(0, \frac{\pi}{2}ight)$.From the first equation: $\log_{10} \sin x + \log_{10} \cos x = -1$$\Rightarrow \log_{10}(\sin x \cos x) = -1$$\Rightarrow \sin x \cos x = 10^{-1} = \frac{1}{10} \quad ....(1)$From the second equation: $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) = \frac{1}{2} \log_{10} \left(\frac{n}{10}ight) = \log_{10} \sqrt{\frac{n}{10}}$$\Rightarrow \sin x + \cos x = \sqrt{\frac{n}{10}}$Squaring both sides:$(\sin x + \cos x)^2 = \frac{n}{10}$$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{n}{10}$$1 + 2 \left(\frac{1}{10}ight) = \frac{n}{10}$$1 + \frac{1}{5} = \frac{n}{10}$$\frac{6}{5} = \frac{n}{10}$$n = \frac{6 	imes 10}{5} = 12$.

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

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