Given that (((tan α)/(sin θ))−((tan β)/(tan θ)))^2 =tan^2 α−tan^2 β, prove that cos θ=((tan β)/(tan α ))


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Question Number 143390 by ZiYangLee last updated on 13/Jun/21

$Given that {(\frac{\tan α}{\sin θ} - \frac{\tan β}{\tan θ})}^{2} = \tan^{2} α - \tan^{2} β,$ $prove that \cos θ = \frac{\tan β}{\tan α}$
	Answered by mindispower last updated on 13/Jun/21

	$\Leftrightarrow {(\frac{1}{s i n (θ)} - \frac{1}{t g (θ)} . \frac{t g (β)}{t g (α)})}^{2} = 1 - \frac{{t g}^{2} (β)}{{t g}^{2} (α)}$ $l e t X = \frac{t g (β)}{t g (α)}$ ${(\frac{1}{s i n (θ)} - \frac{X}{t g (θ)})}^{2} = 1 - X^{2}$ $\Leftrightarrow X^{2} (1 + \frac{1}{{t g}^{2} (θ)}) - 2 \frac{c o s (θ)}{{s i n}^{2} (θ)} X + \frac{{c o s}^{2} (θ)}{{s i n}^{2} (θ)} = 0$ $X^{2} - 2 c o s (θ) X + {c o s}^{2} (θ) = 0$ ${(X - c o s (θ))}^{2} = 0 \Rightarrow X = c o s (θ)$
	Commented by ZiYangLee last updated on 14/Jun/21

	$t h a n k y o u .$
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