If for $x \in\left(0, \frac{\pi}{2}\right), \log _{10} \sin x+\log _{10} \cos x=-1$ and $\log _{10}(\sin x+\cos x)=\frac{1}{2}\left(\log _{10}{ }^{n}-1\right), n>0$, then the value of $n$ is equal to :

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Trigonometric Ratios and Identities

If for

x \in (0, \frac{π}{2}), lo g_{10} sin x + lo g_{10} cos x = - 1

and

lo g_{10} (sin x + cos x) = \frac{1}{2} (lo g_{10}^{n} - 1), n > 0

, then the value of

n

is equal to :

Answer: (B)

From

lo g_{10} (sin x cos x) = - 1

, we get

sin x cos x = \frac{1}{10}

. Let

s = sin x + cos x

. Then

s^{2} = 1 + 2 sin x cos x = 1 + \frac{1}{5} = \frac{6}{5}

, so

s = \frac{6}{5}

. From the second equation:

lo g_{10} s = \frac{1}{2} (lo g_{10} n - 1)

, so

lo g_{10} s = \frac{1}{2} lo g_{10} \frac{n}{10}

, giving

s^{2} = \frac{n}{10}

. Thus

\frac{6}{5} = \frac{n}{10}

, so

n = 12

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option