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Question Number 29413 by math solver last updated on 08/Feb/18

Commented by math solver last updated on 08/Feb/18

$q . 3 ?$
	Answered by beh.i83417@gmail.com last updated on 09/Feb/18
	$x^2 −2x.cosα+cos^2 α+sin^2 α=0 y^2 −2y.cosβ+cos^2 β+sin^2 β=0 (x−cosα)^2 =i^2 sin^2 α⇒x=cosα±i.sinα (y−cosβ)^2 =i^2 sin^2 β⇒y=cosβ±i.sinβ (±ve): { ((x=cosα+i.sinα)),(((1/x)=cosα−i.sinα)) :}⇒ { ((cosα=(1/2)(x+(1/x)))),((sinα=(1/(2i))(x−(1/x)))) :} cos(α+β)=cosα.cosβ−sinα.sinβ cosα.cosβ=(1/4).(((x^2 +1)(y^2 +1))/(xy)) sinα.sinβ=((−1)/4).(((x^2 −1)(y^2 −1))/(xy)) 4cos(α+β)=((x^2 .y^2 +x^2 +y^2 +1+x^2 .y^2 −x^2 −y^2 +1)/(xy)) ⇒2cos(α+β)=((x^2 y^2 +1)/(xy))=xy+(1/(xy)) .■$
	$x^{2} - 2 x . c o s α + {c o s}^{2} α + {s i n}^{2} α = 0$ $y^{2} - 2 y . c o s β + {c o s}^{2} β + {s i n}^{2} β = 0$ ${(x - c o s α)}^{2} = i^{2} {s i n}^{2} α \Rightarrow x = c o s α \pm i . s i n α$ ${(y - c o s β)}^{2} = i^{2} {s i n}^{2} β \Rightarrow y = c o s β \pm i . s i n β$ $(\pm v e) :$ ${\begin{cases} x = c o s α + i . s i n α \\ \frac{1}{x} = c o s α - i . s i n α \end{cases} \Rightarrow {\begin{cases} c o s α = \frac{1}{2} (x + \frac{1}{x}) \\ s i n α = \frac{1}{2 i} (x - \frac{1}{x}) \end{cases}$ $c o s (α + β) = c o s α . c o s β - s i n α . s i n β$ $c o s α . c o s β = \frac{1}{4} . \frac{(x^{2} + 1) (y^{2} + 1)}{x y}$ $s i n α . s i n β = \frac{- 1}{4} . \frac{(x^{2} - 1) (y^{2} - 1)}{x y}$ $4 c o s (α + β) = \frac{x^{2} . y^{2} + x^{2} + y^{2} + 1 + x^{2} . y^{2} - x^{2} - y^{2} + 1}{x y}$ $\Rightarrow 2 c o s (α + β) = \frac{x^{2} y^{2} + 1}{x y} = x y + \frac{1}{x y} . ◼$
	Commented by math solver last updated on 09/Feb/18

	$s i r, a l t h o u g h y o u r m e t h o d i s r i g h t$ $b u t a n s . g i v e n i n b o o k i s o p t i o n (c) .$
	Commented by beh.i83417@gmail.com last updated on 09/Feb/18

	$y o u a r e r i g h t s i r . i t i s n o w c o r r e c t e d .$
	Commented by math solver last updated on 09/Feb/18

	$t h a n k y o u s i r .$
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