BeMath-dx-1-Given-x-1-x-2-y-1-y-2-1-find-x-y-2-

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