Question

#Quality Question
@Logrithm and Trignometry
■》Last 2nd of the day.

⟼The smallest positive AC satisfying the equation
$\sf log_{cos x}(sinx) + log_{sinx}(cosx) = 2 \: is$
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Answer 1

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▪Answer :-

$⟼\green{\large \bf x = \frac{\pi}{4} }$

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▪Given :-

$⟼\bf log_{cos x}(sinx) + log_{sinx}(cosx) = 2$

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▪To Calculate :-

⟼ Smallest positive x satisfying the above equation.

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▪Main Formulae Used :-

$\large \bigstar1 )\: \: \bf log_{n}(m) = \frac{log(m)}{ log(n) } \\ \\ \large \bigstar2) \: \: \bf log_{n}(m) = p \\ \\ \large \bf\implies m = {n}^{p}$

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▪Solution :-

⟼Let

$\sf log_{cos x}(sinx) = y \\ \\ \implies \sf log_{sinx}(cosx) = \frac{1}{y} \\ \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ using \: \: \bigstar1 \}$

So,

$\sf log_{cos x}(sinx) + log_{sinx}(cosx) = 2 \\ \\ \implies \sf y + \frac{1}{y} = 2 \\ \\ \implies \sf {y}^{2} + 1 = 2y \\ \\ \implies \sf {y}^{2} - 2y + 1 = 0 \\ \\ \implies \sf {(y - 1)}^{2} = 0 \\ \\ \bf \implies y = 1$

Hence,

$\sf log_{cos x}(sinx) = 1 \\ \\ \implies \sf sinx = (cosx) {}^{1} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{using \: \bigstar2 \} \\ \\ \implies \sf sinx = cosx \\ \\ \implies \sf \frac{sinx}{cosx} = 1 \\ \\ \implies \sf tanx = 1 \\ \\ \implies \huge\colorbox{skyblue}{\underline{ \boxed{ \bf x = \frac{\pi}{4} }}}$

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Answer 2

Let

$log_{cos x}(sinx) = p \\ \\ \implies \sf log_{sinx}(cosx) = \frac{1}{p} \\ (\bf\because\bf log_{b}(a) =\frac{log(a)}{log(b)})$

Now we can say that

$log_{cos x}(sinx) + log_{sinx}(cosx) = 2 \\ \\ ⟼ p+ \frac{1}{p} = 2 \\ \\ ⟼ {p}^{2} + 1 = 2p \\ \\ ⟼ {(p - 1)}^{2} = 0 \\ \\⟼ \boxed{\mathfrak{ p = 1}}$

Which means

$log_{cos x}(sinx) = 1 \\ \\ ⟼ sinx = (cosx) {}^{1} \\ (\bf\because log_b(a)=c⟼a=b^c)\\ \\ ⟼ sinx = cosx \\ \\⟼ tanx = 1 \\ \\ ⟼\Large{\boxed{ \boxed{\pink{ \bf x = \frac{\pi}{4} }}}}$