Question

A challenge to all math experts...

Answer 1

Answer:

0

Step-by-step explanation:

Given

• (x + $\sqrt{1 + x^2}$)(y + $\sqrt{1 + y^2}$) = 1
• (x+y)² = ?

Solution

Let's make substitution as:

• (x + $\sqrt{1 + x^2}$) = m

then

• (y + $\sqrt{1 + y^2}$) = 1/m

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

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

And

• (y + $\sqrt{1 + y^2}$) = 1/m
• $\sqrt{1 + y^2}$ = 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