If \[ \tan x=\frac{b}{a} \] Find the Value of \[ \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}} \]

Solution:

\[ \sqrt{\frac{a+b}{a-b}} + \sqrt{\frac{a-b}{a+b}} \]

Taking LCM,

\[ = \frac{ (a+b)+(a-b) } { \sqrt{(a+b)(a-b)} } \]

\[ = \frac{2a}{\sqrt{a^2-b^2}} \]

Since \[ \tan x=\frac{b}{a} \]

\[ 1-\tan^2 x = 1-\frac{b^2}{a^2} = \frac{a^2-b^2}{a^2} \]

\[ \sqrt{a^2-b^2} = a\sqrt{1-\tan^2 x} \]

Therefore,

\[ = \frac{2a}{a\sqrt{1-\tan^2 x}} \]

\[ = \frac{2}{\sqrt{1-\tan^2 x}} \]