JEE Main 2021 (Online) 16th March Morning ShiftQuestion 1
If for $x\in\left(0,\dfrac{\pi}{2}\right)$, $\log_{10}\sin x+\log_{10}\cos x=-1$ and $\log_{10}(\sin x+\cos x)=\dfrac{1}{2}\left(\log_{10}n-1\right)$, $n>0$, then the value of $n$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. The first condition gives $\sin x\cos x$; squaring $\sin x+\cos x$ uses it. Formulas used. (1) $\log p+\log q=\log pq$; $(\sin x+\cos x)^2=1+2\sin x\cos x$Step 1 - Product. $\sin x\cos x=\frac{1}{10}$. Step 2 - Square. $(\sin x+\cos x)^2=1+\frac15=\frac65$, so $\log_{10}(\sin x+\cos x)=\frac12\log_{10}\frac{12}{10}=\frac12(\log_{10}12-1)$. Step 3 - Value. $n=12$. Answer: (B)