If for x ∈ (0, (π)/(2)), _ 10 x + _ 10 x = - 1 and _ 10 ( x + x) =...

Given $x \in \left(0, \dfrac{\pi}{2}\right)$

$\log_{10}\sin x + \log_{10}\cos x = - 1$

$⇒ \log_{10}\sin x . \cos x = - 1$

$⇒ \sin x . \cos x = \dfrac{1}{10}$ $. . . ( 1 )$

$\log_{10}\left(\sin x + \cos x\right) = \dfrac{1}{2}\left(\log_{10}n - 1\right)$

$⇒ \sin x + \cos x = {10}^{\left(\log_{10}\sqrt{n} - \dfrac{1}{2}\right)} = {10}^{\left(\log_{10}\sqrt{n} - \log_{10}\sqrt{10}\right)} = \sqrt{\dfrac{n}{10}}$

By squaring we get,

$\sin^{2}x + \cos^{2}x + 2\sin x \cos x = \dfrac{n}{10}$

$1 + 2\sin x . \cos x = \dfrac{n}{10}$

$⇒ 1 + \dfrac{1}{5} = \dfrac{n}{10} ⇒ n = 12$