oolve cos 2x .tan (((7π)/(19)))=tan (((17π)/(23)))+tan (((6π)/(23)))+tan (((12π)/(19)))


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Question Number 193502 by horsebrand11 last updated on 15/Jun/23

$\underset{―}{o} olve$ $\cos 2 x . \tan (\frac{7 π}{19}) = \tan (\frac{17 π}{23}) + \tan (\frac{6 π}{23}) + \tan (\frac{12 π}{19})$
	Answered by cortano12 last updated on 15/Jun/23

	$If α + β = π then {\begin{cases} \tan α + \tan β = 0 \\ \tan α \cot β = - 1 \end{cases}$ $so \cos 2 x \tan (\frac{7 π}{19}) = 0 + \tan (\frac{12 π}{19})$ $\Leftrightarrow \cos 2 x = - 1$ $\Leftrightarrow \cos 2 x = \cos π$ $\Leftrightarrow x = \pm \frac{π}{2} + k . π$
	Commented by BaliramKumar last updated on 15/Jun/23

	$\sin 2 x \to \cos 2 x$
	Answered by aba last updated on 15/Jun/23

	$\Rightarrow \cos 2 x . tg \frac{7 π}{19} = tg (π - \frac{6 π}{23}) + tg (\frac{6 π}{23}) + tg (π - \frac{7 π}{19})$ $\Rightarrow \cos 2 x . tg \frac{7 π}{19} = - tg \frac{6 π}{23} + tg \frac{6 π}{23} - tg \frac{7 π}{19}$ $\Rightarrow \cos 2 x = - 1 = \cos (π)$ $\Rightarrow {\begin{cases} 2 x \equiv π [2 π] \\ 2 x \equiv - π [2 π] \end{cases} \Rightarrow {\begin{cases} x \equiv \frac{1}{2} π [2 π] = \frac{1}{2} π (2 k + 1) \\ x \equiv - \frac{1}{2} π [2 π] = \frac{1}{2} π (2 k - 1) \end{cases} k \in Z$
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