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Question
If \[\tan x = \frac{b}{a}\] , then find the values of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\].
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Solution
\[\tan x = \frac{b}{a}\]
\[\text{ Now }, \sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\]
\[ = \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}}\]
\[ = \sqrt{\frac{1 + \tan x}{1 - \tan x}} + \sqrt{\frac{1 - \tan x}{1 + \tan x}}\]
\[ = \frac{\tan x + 1 + 1 - \tan x}{\sqrt{1 - \tan^2 x}}\]
\[ = \frac{2}{\sqrt{1 - \tan^2 x}}\]
\[ = \frac{2\cos x}{\sqrt{\cos^2 x - \sin^2 x}}\]
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