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Question
If \[\text{ tan } x = \frac{b}{a}\] , then find the value of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\] .
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Solution
Given: \[\text{ tan } x = \frac{b}{a}\]
\[\therefore \sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\]
\[ = \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}}\]
\[ = \sqrt{\frac{1 + \text{ tan } x}{1 - \text{ tan } x}} + \sqrt{\frac{1 - \text{ tan } x}{1 + \text{ tan } x}}\]
\[ = \sqrt{\frac{1 + \frac{\text{ sin } x}{\text{ cos } x}}{1 - \frac{\text{ sin } x}{\text{ cos } x}}} + \sqrt{\frac{1 - \frac{\text{ sin } x}{\text{ cos } x}}{1 + \frac{\text{ sin } x}{\text{ cos } x}}}\]
\[= \sqrt{\frac{\text{ cos } x + \text{ sin } x}{\text{ cos } x - \text{ sin } x}} + \sqrt{\frac{\text{ cos } x - \text{ sin } x}{\text{ cos } x + \text{ sin } x}}\]
\[ = \frac{\text{ cos } x + \text{ sin } x + \text{ cos } x - \text{ sin } x}{\sqrt{\left( \text{ cos } x - \text{ sin } x \right)\left( \text{ cos } x + \text{ sin } x \right)}}\]
\[ = \frac{2\text{ cos } x}{\sqrt{\text{ cos } ^2 x - \sin^2 x}}\]
\[ = \frac{2\text{ cos } x}{\sqrt{\cos2x}}\]
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