Trigonometric Ratios and Identities
Logarithms and Trigonometry
JEE Main 2021
Grade 11
Question:
If $\log_{10} \sin x + \log_{10} \cos x = -1$ and $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$, $x \in (0, \frac{\pi}{2})$, then $n$ equals:
(1) $9$
(2) $10$
(3) $11$
(4) $12$
Step-by-Step Solution
Key Concept: The first equation gives $\sin x \cos x = \frac{1}{10}$, so $\sin 2x = \frac{1}{5}$. Then $(\sin x + \cos x)^2 = 1 + \sin 2x$. Substitute into the second equation and solve for $n$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (4)