Step-by-step explanation:
Given that: [√(1+x)/{√(1+x) - √(1-x)}] - [(1-x)/{√(1-x²) + x -1}]
Here, we see that in first fraction the denominator is √(1+x)-√(1-x) , as we know that the rational factor √(a+b)-√(a-b) is √(a+b)+(a-b). Therefore, the rationalising factor of √(1+x)-√(1-x) is √(1+x)+√(1-x). On rationalising the denominator them.
= [√(1+x)/{√(1+x)-√(1-x)}] * [{√(1+x)+√(1-x)}/{√(1+x)+√(1-x)}] - [(1-x)/{√(1-x²) + x -1}]
= [{√(1+x)(√(1+x)+√(1-x)}/{√(1+x)-√(1-x) - √(1+x)+√(1-x)}] - [(1-x)/{√(1-x²) + x -1}]
Now, comparing the first fraction denominator with (a-b)(a+b), we get
- a = √(1+x)
- b = √(1-x)
- a = √(1+x)
- b = √(1-x)
Using identity (a-b)(a+b) = a² - b², we get
= [{√(1+x)(√(1+x)+√(1-x)}/{√(1+x)² - √(1-x)²}] - [(1-x)/{√(1-x²) + x -1}]
= [{√(1+x)(√(1+x)+√(1-x)}/{(1+x) - (1-x)}] - [(1-x)/{√(1-x²) + x -1}]
Now, multiply the numerator on both brackets.
= [{√(1+x) * √(1+x) + √(1+x) * √(1-x)}/{(1+x) - (1-x)}] - [(1-x)/{√(1-x²) + x -1}]
Comparing the first fraction numerator with (a+b)(a+b) , we get
Using identity (a+b)(a+b) = (a+b)², we get
= [{√(1+x)²+ √(1+x) * √((1+x)(1-x))}/{(1+x) - (1-x)}] - [(1-x)/{√(1-x²) + x -1}]
Cancel out the first fraction denominator numbers 1 and -1 to get 0.
= [{(1+x)+√(1-x²)}/(x+x+0)] - [(1-x)/{√(1-x²) + x -1}]
= [{(1+x)+√(1-x²)}/(x+x)] - [(1-x)/{√(1-x²) + x -1}]
= [{(1+x)+√(1-x²)}/2x] - [(1-x)/{√(1-x²) + x -1}]
= [{1+x+√(1-x²)}{√(1-x²)x-1}-{(1-x)(2x)}]/[2x{√(1-x²)+x-1}]
= [√(1-x²) + x-1 + x√(1-x²) + x² - x + {√(1-x²)}² + x√(1-x²)-√(1-x²) - 2x + 2x²]/[2x{√(1-x²)+x-1}]
= {(√(1-x²) 2x -1 + 3x² + 2x√(1-x²) + 1-x² - √(1-x²)}/[2x{√(1-x²)+x-1}]
Cancel out √(1-x) and -√(1-x) in numerator.
= {-2x - 1 + 2x√(1-x²) + 3x² + 1 - x²}/[2x{√(1-x²)}+x-1}]
Cancel out -1 and 1 in numerator to get 0.
= [2x√(1-x²) - 2x + 2x²}/[2x{√(1-x²)+x-1}]
= {2x{√(1-x²)} + x - 1}]/[2x{√(1-x²)}+x-1}
= 1
<u>Answer</u><u>:</u> Hence, the simplified form of the expression [√(1+x)/{√(1+x) - √(1-x)}] - [(1-x)/{√(1-x²) + x -1}] is 1.
Please let me know if you have any other questions.
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