If `"log"_(10) sin x+"log"_(10)cos x=-1 and "log"_(10)(sin x+cosx)=(("log"_(10)n)n-1)/(2)` then the value of n is ____________

Answer

12

Explanation

Given `log_(10)((sin2x )/( 2))= -1 `
or `( sin 2x )/(2)= (1)/(10)`
or `sin 2x = (1)/(5)" " `…(i)
Also `log_(10) ( sin x + cos x ) = ( log_(10)((n)/(10)))/(2)`
or `log_(10)(sin x + cosx )^(2) = log_(10) ((n)/(10))`
or `1 + sin 2x = (n)/(10)`
or `1 + (1)/(5) = (n)/(10)`
or `(6)/(5) = (n)/(10)`
`therefore n = 12`