If for x∈ (0, π 2), log_10 sinx+ log_10 cosx=-1 and log_10( sinx+ cosx)= 1 2( log_10n-1),n>0, then the value of n is equal to :

Given $x\in (0,\frac{\pi }{2})$x∈(0,2π​)

$${\mathrm{log}}_{10}\mathrm{sin}x+{\mathrm{log}}_{10}\mathrm{cos}x=-1$$log10​sinx+log10​cosx=−1

$$\Rightarrow {\mathrm{log}}_{10}\mathrm{sin}x.\mathrm{cos}x=-1$$⇒log10​sinx.cosx=−1

$\Rightarrow \mathrm{sin}x.\mathrm{cos}x=\frac{1}{10}$⇒sinx.cosx=101​ $...(1)$...(1)

$${\mathrm{log}}_{10}(\mathrm{sin}x+\mathrm{cos}x)=\frac{1}{2}({\mathrm{log}}_{10}n-1)$$log10​(sinx+cosx)=21​(log10​n−1)

$$\Rightarrow \mathrm{sin}x+\mathrm{cos}x={10}^{({\mathrm{log}}_{10}\sqrt{n}-\frac{1}{2})}={10}^{({\mathrm{log}}_{10}\sqrt{n}-{\mathrm{log}}_{10}\sqrt{10})}=\sqrt{\frac{n}{10}}$$⇒sinx+cosx=10(log10​n​−21​)=10(log10​n​−log10​10​)=10n​​

By squaring we get,

$${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x+2\mathrm{sin}x\mathrm{cos}x=\frac{n}{10}$$sin2x+cos2x+2sinxcosx=10n​

$$1+2\mathrm{sin}x.\mathrm{cos}x=\frac{n}{10}$$1+2sinx.cosx=10n​

$$\Rightarrow 1+\frac{1}{5}=\frac{n}{10}\Rightarrow n=12$$⇒1+51​=10n​⇒n=12