If for x ∈ 0 , π 2 , log 10 sin x + log 10 cos x = - 1 and log 10 sin x + cos x = 1 2 log 10 n - 1 , n > 0 , then the value of n is equal to :

Step-by-step solution

Given x ∈ 0 , π 2 log 10 sin x + log 10 cos x = - 1 ⇒ log 10 sin x . cos x = - 1 ⇒ sin x . cos x = 1 10 . . . ( 1 ) log 10 sin x + cos x = 1 2 log 10 n - 1 ⇒ sin x + cos x = 10 log 10 n - 1 2 = 10 log 10 n - log 10 10 = n 10 By squaring we get, sin 2 x + cos 2 x + 2 sin x cos x = n 10 1 + 2 sin x . cos x = n 10 ⇒ 1 + 1 5 = n 10 ⇒ n = 12