Exams Joint Entrance Examination Mathematics Trigonometric Ratios and Identities Question 288 Question 288 of 298 Medium MCQ
If for x \[\in\] \[\left( {0,{\pi \over 2}} \right)\], log10sinx + log10cosx = -1 and log10(sinx + cosx) = \[{1 \over 2}(log10 n - 1), n > 0, then the value of n is equal to :
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log 10 sin x + log 10 cos x = − 1 , x ∈ ( 0 , π / 2 ) log 10 ( sin x cos x ) = − 1 ⇒ sin x cos x = 10 − 1 = 1 10 log 10 ( sin x + cos x ) = 1 2 ( log 10 n − 1 ) , n > 0 2 log 10 ( sin x + cos x ) = ( log 10 n − log 10 10 ) ⇒ log 10 ( sin x + cos x ) 2 = log 10 ( n 10 ) ⇒ ( sin x + cos x ) 2 = n 10 ⇒ sin 2 x + cos 2 x + 2 sin x cos x = n 10 ⇒ 1 + 2 ( 1 10 ) = n 10 ⇒ 12 10 = n 10 ∴ n = 12 \begin{aligned}
& \log _{10} \sin x+\log _{10} \cos x=-1, x \in(0, \pi / 2) \\\\
& \log _{10}(\sin x \cos x)=-1 \\\\
& \Rightarrow \sin x \cos x=10^{-1}= {1 \over {10}} \\\\
& \log _{10}(\sin x+\cos x)={1 \over {2}}\left(\log _{10} n-1\right), n>0 \\\\
& 2 \log _{10}(\sin x+\cos x)=\left(\log _{10} n-\log _{10} 10\right) \\\\
& \Rightarrow \log _{10}(\sin x+\cos x)^2=\log _{10}({n \over {10}}) \\\\
& \Rightarrow (\sin x+\cos x)^2={n \over {10}} \\\\
& \Rightarrow \sin ^2 x+\cos ^2 x+2 \sin x \cos x=\frac{n}{10} \\\\
& \Rightarrow 1+2({1 \over {10}})={n \over {10}} \Rightarrow {12 \over {10}}={n \over {10}} \\\\
& \therefore n=12
\end{aligned} log 10 sin x + log 10 cos x = − 1 , x ∈ ( 0 , π /2 ) log 10 ( sin x cos x ) = − 1 ⇒ sin x cos x = 1 0 − 1 = 10 1 log 10 ( sin x + cos x ) = 2 1 ( log 10 n − 1 ) , n > 0 2 log 10 ( sin x + cos x ) = ( log 10 n − log 10 10 ) ⇒ log 10 ( sin x + cos x ) 2 = log 10 ( 10 n ) ⇒ ( sin x + cos x ) 2 = 10 n ⇒ sin 2 x + cos 2 x + 2 sin x cos x = 10 n ⇒ 1 + 2 ( 10 1 ) = 10 n ⇒ 10 12 = 10 n ∴ n = 12
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