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A challenge to all math expertsFollow me if you love such challenging questions as I keep on asking such questions to check the mindsets of the mathematicians r class=

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mhanifa

Answer:

0

Step-by-step explanation:

Given

  • (x + [tex]\sqrt{1 + x^2}[/tex])(y + [tex]\sqrt{1 + y^2}[/tex]) = 1
  • (x+y)² = ?

Solution

Let's make substitution as:

  • (x + [tex]\sqrt{1 + x^2}[/tex]) = m

then

  • (y + [tex]\sqrt{1 + y^2}[/tex]) = 1/m

Solving the first equation for x and the second one for y:

  • (x + [tex]\sqrt{1 + x^2}[/tex]) = m
  • [tex]\sqrt{1 + x^2}[/tex] = m - x
  • 1 + x² = (m - x)²
  • 1 + x² = m² - 2mx + x²
  • 2mx = m² - 1
  • x = (m² - 1)/2m

And

  • (y + [tex]\sqrt{1 + y^2}[/tex]) = 1/m
  • [tex]\sqrt{1 + y^2}[/tex] = 1/m - y
  • 1 + y² = 1/m² -2y/m + y²
  • 1 = 1/m² - 2y/m
  • m² = 1 - 2my
  • 2my = 1 - m²
  • y = (1 - m²)/2m
  • y = - (m² - 1)/2m

Now, sum of x and y:

  • x + y =  (m² - 1)/2m - (m² - 1)/2m = 0

Therefore:

  • (x+y)² = 0

Answer is zero

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