[√(1 +x)+√(1 −x)] [√(1 +x)−√(1 −x)] I'd rather change these into variables: a = √(1 +x) b = √(1 −x) (a+b) (a-b) = a² - b² When √1 is squared, it's valued as the number itself. (√1)² = 1 Answer: (1+x) - (1-x) 1 + x - 1 + x 2x So, [√(1 +x)+√(1 −x)] [√(1 +x)−√(1 −x)] = 2x [√(1 +x)−√(1 −x)] / 2x = 1 / [√(1 +x)+√(1 −x) ] 1/ [√(1+x) + √(1-x)] = [√(1+x) - √(1-x)] / 2x Equals with question request. ====================== part II (1/ [√(1+x) + √(1-x)])² = [√(1+x) - √(1-x)] / 2x 1/ [√(1+x) + √(1-x)]² = [√(1+x) - √(1-x)] / 2x 1/ (a+b)² = (a-b) / 2x Use ad x bc method. 2x =
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