If for x∈(0, π2),log10sinx+log10cosx=−1 and log10(sinx+cosx)=12(log10n−1), n>0 then the value of n is equal to:

sinx.cosx=110 (sinx+cosx)2=n101+sin2x=n10 1+15=n10 n=(65)×10=12

If for x∈(0, π2),log10sinx+log10cosx=−1 and log10(sinx+cosx)=12(log10n−1), n>0 then the value of n is equal to:

sinx.cosx=110 (sinx+cosx)2=n101+sin2x=n10 1+15=n10 n=(65)×10=12

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If for $x \in (0, \frac{π}{2}), \log_{10} \sin x + \log_{10} \cos x = - 1$ 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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$\sin x . \cos x = \frac{1}{10} {(\sin x + \cos x)}^{2} = \frac{n}{10}$

$1 + \sin 2 x = \frac{n}{10} 1 + \frac{1}{5} = \frac{n}{10} n = (\frac{6}{5}) \times 10 = 12$

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If for x∈(0, π2),log10sinx+log10cosx=−1 and log10(sinx+cosx)=12(log10n−1), n>0 then the value of n is equal to:

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