Using properties of proportion, solve each of the following for $x$ :
(i) $\frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \frac{a}{b}$ .

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Solution

Calculate the value of $x$ :
If $\frac{a}{b} = \frac{c}{d}$ , then the property of componendo and dividendo states that $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ .
The given proportion can be simplified as,
$\frac{\sqrt{1 + x} + \sqrt{1 - x} + \sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x} - (\sqrt{1 + x} - \sqrt{1 - x})} = \frac{a + b}{a - b} (b y c o m p o n e n d o a n d d i v i d e n d o) \Rightarrow \frac{2 \sqrt{1 + x}}{2 \sqrt{1 - x}} = \frac{a + b}{a - b} \Rightarrow \frac{\sqrt{1 + x}}{\sqrt{1 - x}} = \frac{a + b}{a - b} \Rightarrow \frac{1 + x}{1 - x} = \frac{{(a + b)}^{2}}{{(a - b)}^{2}} (S q u a r i n g b o t h s i d e s) \Rightarrow \frac{1 + x + 1 - x}{1 + x - 1 + x} = \frac{{(a + b)}^{2} + {(a - b)}^{2}}{{{(a + b)}^{2} - (a - b)}^{2}} (b y c o m p o n e n d o a n d d i v i d e n d o) \Rightarrow \frac{2}{2 x} = \frac{2 (a^{2} + b^{2})}{4 a b} (U \sin g t h e f o r m u l a e {(a - b)}^{2} = a^{2} - 2 a b + b^{2} a n d {(a + b)}^{2} = a^{2} + 2 a b + b^{2}) \Rightarrow \frac{1}{x} = \frac{(a^{2} + b^{2})}{2 a b} \Rightarrow x = \frac{2 a b}{(a^{2} + b^{2})} (T a k i n g t h e r e c i p r o c a l)$
Final Answer: The value of $x = \frac{2 a b}{a^{2} + b^{2}}$ .

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