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JEE Main 2020
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

If for x \in (0,π2)\left( {0,{\pi \over 2}} \right), log 10 sinx + log 10 cosx = -1 and log 10 (sinx + cosx) = 12{1 \over 2}(log 10 n - 1), n > 0, then the value of n is equal to :

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Solution

log10sinx+log10cosx=1,x(0,π/2)log10(sinxcosx)=1sinxcosx=101=110log10(sinx+cosx)=12(log10n1),n>02log10(sinx+cosx)=(log10nlog1010)log10(sinx+cosx)2=log10(n10)(sinx+cosx)2=n10sin2x+cos2x+2sinxcosx=n101+2(110)=n101210=n10n=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}

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