Question

If for `x∈(0, π/2), log_10sinx + log_10cosx` = –1 and `log_10(sinx + cosx) = 1/2(log_10n - 1), n > 0`, then the value of n is equal to ______.

पर्याय

• 16

• 20

• 12

• 9

Answer 1

If for `x∈(0, π/2), log_10sinx + log_10cosx` = –1 and `log_10(sinx + cosx) = 1/2(log_10n - 1), n > 0`, then the value of n is equal to 12.

Explanation:

Given `log_10sinx + log_10cosx = -1, x∈(0, π/2)`

log₁₀(sinx.cosx) = –1

sinx.cosx = (10)^–1

sinx.cosx = `1/10`  ...(i)

Given, `log_10(sinx + cosx) = 1/2(log_10n - 1), n > 0`

Using log_a a = 1, we can write

`2log_10(sinx + cosx) = 1/2(log_10 n - log_10 10)`

`log_10(sinx + cosx)^2 = log_10(n/10)`

⇒ `(sinx + cosx)^2 = n/10`

⇒ `sin^2x + cos^2x + 2sinxcosx = n/10`

⇒ `1 + 2sinx.cosx = n/10`  ...(ii)

Now using equation (i), we can write

⇒ `1 + 2(1/10) = n/10`

⇒ `1 + 1/5 = n/10`

⇒ `6/5 = n/10`

⇒ n = 12