Given x∈(0,2π)
log10sinx+log10cosx=−1
⇒log10sinx.cosx=−1
⇒sinx.cosx=101 ...(1)
log10(sinx+cosx)=21(log10n−1)
⇒sinx+cosx=10(log10n−21)=10(log10n−log1010)=10n
By squaring we get,
sin2x+cos2x+2sinxcosx=10n
1+2sinx.cosx=10n
⇒1+51=10n⇒n=12