Write an article in about 100-120 words expressing your ideas on how online learning is here to stay and how it can be used to its optimum
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▪Answer :-
\begin{gathered} ⟼\green{\large \bf x = \frac{\pi}{4} } \\ \end{gathered}⟼x=4π
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▪Given :-
⟼\bf log_{cos x}(sinx) + log_{sinx}(cosx) = 2⟼logcosx(sinx)+logsinx(cosx)=2
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▪To Calculate :-
⟼ Smallest positive x satisfying the above equation.
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▪Main Formulae Used :-
\begin{gathered} \large \bigstar1 )\: \: \bf log_{n}(m) = \frac{log(m)}{ log(n) } \\ \\ \large \bigstar2) \: \: \bf log_{n}(m) = p \\ \\ \large \bf\implies m = {n}^{p} \end{gathered}★1)logn(m)=log(n)log(m)★2)logn(m)=p⟹m=np
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▪Solution :-
⟼Let
\begin{gathered}\sf log_{cos x}(sinx) = y \\ \\ \implies \sf log_{sinx}(cosx) = \frac{1}{y} \\ \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ using \: \: \bigstar1 \}\end{gathered}logcosx(sinx)=y⟹logsinx(cosx)=y1 {using★1}
So,
\begin{gathered}\sf log_{cos x}(sinx) + log_{sinx}(cosx) = 2 \\ \\ \implies \sf y + \frac{1}{y} = 2 \\ \\ \implies \sf {y}^{2} + 1 = 2y \\ \\ \implies \sf {y}^{2} - 2y + 1 = 0 \\ \\ \implies \sf {(y - 1)}^{2} = 0 \\ \\ \bf \implies y = 1\end{gathered}logcosx(sinx)+logsinx(cosx)=2⟹y+y1=2⟹y2+1=2y⟹y2−2y+1=0⟹(y−1)2=0⟹y=1
Hence,
\begin{gathered}\sf log_{cos x}(sinx) = 1 \\ \\ \implies \sf sinx = (cosx) {}^{1} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{using \: \bigstar2 \} \\ \\ \implies \sf sinx = cosx \\ \\ \implies \sf \frac{sinx}{cosx} = 1 \\ \\ \implies \sf tanx = 1 \\ \\ \implies \huge\colorbox{skyblue}{\underline{ \boxed{ \bf x = \frac{\pi}{4} }}}\end{gathered}logcosx(sinx)=1⟹sinx=(cosx)1{using★2}⟹sinx=cosx⟹cosxsinx=1⟹tanx=1⟹ x=4π
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