((∥ BeMath ∥)/(°∫ dx°)) (1) Given (x+(√(1+x^2 )))(y+(√(1+y^2 )))=1 find (x+y)^2


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Question Number 108291 by bemath last updated on 16/Aug/20

$\frac{∥ B e M a t h ∥}{° \int d x °}$ $(1) G i v e n (x + \sqrt{1 + x^{2}}) (y + \sqrt{1 + y^{2}}) = 1$ $f i n d {(x + y)}^{2}$
	Answered by john santu last updated on 16/Aug/20

	$\frac{⇈ J S ⇈}{♡}$ $\Leftrightarrow s e t u = x + \sqrt{1 + x^{2}}$ $\Rightarrow u . (y + \sqrt{1 + y^{2}}) = 1$ $y + \sqrt{1 + y^{2}} = \frac{1}{u} \Rightarrow \sqrt{1 + y^{2}} = \frac{1}{u} - y$ $1 + y^{2} = {(\frac{1}{u} - y)}^{2}$ $1 + y^{2} = \frac{1}{u^{2}} - \frac{2 y}{u} + y^{2}$ $1 = \frac{1}{u^{2}} - \frac{2 y}{u}; \frac{2 y}{u} = \frac{1}{u^{2}} - 1$ $2 y = \frac{1}{u} - u \Rightarrow y = \frac{1}{2} (\frac{1}{u} - u)$ $\frac{1}{u} = \frac{1}{x + \sqrt{1 + x^{2}}} \times \frac{x - \sqrt{1 + x^{2}}}{x - \sqrt{1 + x^{2}}}$ $\frac{1}{u} = \frac{x - \sqrt{1 + x^{2}}}{x^{2} - (1 + x^{2})} = \sqrt{1 + x^{2}} - x$ $y = \frac{1}{2} ((\sqrt{1 + x^{2}} - x) - (x + \sqrt{1 + x^{2}}))$ $y = \frac{1}{2} (- 2 x) = - x$ $t h e r e f o r e {(x + y)}^{2} = 0^{2} = 0$
	Commented by bemath last updated on 16/Aug/20

	$n i c e!!$
	Commented by udaythool last updated on 17/Aug/20

	$From given condition we have$ ${(x + \sqrt{1 + x^{2}})}^{2} {(y + \sqrt{1 + y^{2}})}^{2} = 1$ $\Rightarrow {(x + y)}^{2} = - (x \sqrt{1 + x^{2}} + y \sqrt{1 + y^{2}})$ $\Rightarrow {(x + y)}^{2} = 0 ?$
	Commented by udaythool last updated on 17/Aug/20

	$x + \sqrt{1 + x^{2}} = \sqrt{1 + y^{2}} - y$ $\Rightarrow x + y = \sqrt{1 + y^{2}} - \sqrt{1 + x^{2}}$ $\Rightarrow x^{2} + y^{2} + 2 x y = 1 + y^{2} + 1 + x^{2} - 2 \sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}}$ $\Rightarrow 1 - x y = \sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}}$ $\Rightarrow - 2 x y = x^{2} + y^{2}$ $\Rightarrow {(x + y)}^{2} = 0$
	Answered by 1549442205PVT last updated on 16/Aug/20

	$(x + \sqrt{1 + x^{2}}) (y + \sqrt{1 + y^{2}}) = 1 (1)$ $Put x = tanA, y = tanB (A, B \in (- \frac{π}{2}; \frac{π}{2}))$ $Since 1 + \tan^{2} θ = \frac{1}{\cos^{2} θ}, we have$ $(1) \Leftrightarrow (tanA + \frac{1}{cosA}) (tanB + \frac{1}{cosB}) = 1$ $\Leftrightarrow (1 + sinA) (1 + sinB) = cosAcosB$ $\Leftrightarrow 1 + sinA + sinB + sinAsinB = cosAcosB (2)$ $Put \tan \frac{A}{2} = m, \tan \frac{B}{2} = n, since$ $\sin θ = \frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} θ} and \cos θ = \frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}, so$ $(2) \Leftrightarrow 1 + \frac{2 m}{1 + m^{2}} + \frac{2 n}{1 + n^{2}} + \frac{4 mn}{(1 + m^{2}) (1 + n^{2})} = \frac{(1 - m^{2}) (1 - n^{2})}{(1 + m^{2}) (1 + n^{2})}$ $\Leftrightarrow (1 + m^{2}) (1 + n^{2}) + 2 m (1 + n^{2}) + 2 n (1 + m^{2}) + 4 mn = 1 - m^{2} - n^{2} + m^{2} n^{2}$ $\Leftrightarrow {2 m}^{2} + {2 n}^{2} + 2 (m + n) + 2 mn (m + n) + 4 mn = 0$ $\Leftrightarrow 2 {(m + n)}^{2} + 2 (m + n) (1 + mn) = 0$ $\Leftrightarrow 2 (m + n) (1 + m + n + mn) = 0$ $\Leftrightarrow 2 (m + n) (1 + m) (1 + n) = 0$ $If 1 + m = 0 \Rightarrow \tan \frac{A}{2} = - 1 \Rightarrow \frac{A}{2} = \frac{- π}{4}$ $\Rightarrow A = \frac{- π}{2} this is contradicrion$ $to the hypothesis, so 1 + m \neq 0$ $Similarly, 1 + n \neq 0 . Hence we must have$ $m + n = 0 \Rightarrow \tan \frac{A}{2} + \tan \frac{B}{2} = 0$ $\Rightarrow \frac{\tan (\frac{A + B}{2})}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = 0 \Rightarrow \tan \frac{A + B}{2} = 0$ $\Rightarrow A + B = k π \Rightarrow x + y = tanA + tanB$ $= \frac{\tan (A + B)}{1 - tanAtanB} = \frac{0}{1 - tanAtanB} = 0$ $\Rightarrow {(x + y)}^{2} = 0 . Thus, {(x + y)}^{2} = 0$ $second way$ $(2) \Leftrightarrow cosAcosB - sinAsinB = 1 + sinA + sinB$ $\Leftrightarrow \cos (A + B) = 1 + 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}$ $\Leftrightarrow 1 - {2 \sin}^{2} \frac{A + B}{2} = 1 + 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}$ $\Leftrightarrow 2 \sin \frac{A + B}{2} (\cos \frac{A - B}{2} + 1) = 0 (3)$ $Since A, B \in (- \frac{π}{2}; \frac{π}{2}), \frac{A - B}{2} \neq π$ $\Rightarrow \cos \frac{A - B}{2} \neq - 1 \Rightarrow \cos \frac{A - B}{2} + 1 \neq 0$ $Hence (3) \Leftrightarrow \sin \frac{A + B}{2} = 0 \Rightarrow \frac{A + B}{2} = 0$ $\Rightarrow A + B = 0 \Rightarrow \tan (A + B) = 0$ $\Rightarrow tanA + tanB = \frac{\tan (A + B)}{1 - tanAtanB} = 0$ $Therefore, {(x + y)}^{2} = {(tanA + tanB)}^{2} = 0$
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