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Question Number 199413 by universe last updated on 03/Nov/23

	Answered by ajfour last updated on 03/Nov/23
	$bcos α=R {Rtan α+((a/b))R}^2 +{((a/b))Rtan α}^2 =R^2 say tan α=t ⇒ t^2 +(a^2 /b^2 )+((2at)/b)+(a^2 /b^2 )t^2 =1 (a^2 +b^2 )t^2 +2abt+a^2 −b^2 =0 t=tan α=(((√(a^2 b^2 +(b^4 −a^4 )))−ab)/(a^2 +b^2 ))$
	$b \cos α = R$ ${R \tan α + (\frac{a}{b}) R}^{2} + {(\frac{a}{b}) R \tan α}^{2} = R^{2}$ $s a y \tan α = t$ $\Rightarrow t^{2} + \frac{a^{2}}{b^{2}} + \frac{2 a t}{b} + \frac{a^{2}}{b^{2}} t^{2} = 1$ $(a^{2} + b^{2}) t^{2} + 2 a b t + a^{2} - b^{2} = 0$ $t = \tan α = \frac{\sqrt{a^{2} b^{2} + (b^{4} - a^{4})} - a b}{a^{2} + b^{2}}$
	Answered by mr W last updated on 03/Nov/23

	$R = b \cos α$ $R^{2} = a^{2} + {(b \sin α)}^{2} - 2 a b \sin α \cos (\frac{π}{2} + \frac{π}{2} - α)$ $b^{2} \cos^{2} α = a^{2} + b^{2} \sin^{2} α + a b \sin 2 α$ $b^{2} \cos 2 α - a b \sin 2 α = a^{2}$ $l e t t = \tan α$ $b^{2} \times \frac{1 - t^{2}}{1 + t^{2}} - a b \times \frac{2 t}{1 + t^{2}} = a^{2}$ $(a^{2} + b^{2}) t^{2} + 2 a b t + a^{2} - b^{2} = 0$ $\Rightarrow t = \frac{\sqrt{a^{2} b^{2} + b^{4} - a^{4}} - a b}{a^{2} + b^{2}} = \tan α$
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