Question

If $\tan \mathit{}x=\frac{b}{a}$, then find the values of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$.

Answer 1

$$\begin{gathered} \tan x=\frac{b}{a} \\ \mathrm{Now},\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} \\ =\sqrt{\frac{1+\frac{b}{a}}{1-\frac{b}{a}}}+\sqrt{\frac{1-\frac{b}{a}}{1+\frac{b}{a}}} \\ =\sqrt{\frac{1+\tan x}{1-\tan x}}+\sqrt{\frac{1-\tan x}{1+\tan x}} \\ =\frac{\tan x+1+1-\tan x}{\sqrt{1-{\tan }^{2}x}} \\ =\frac{2}{\sqrt{1-{\tan }^{2}x}} \\ =\frac{2\cos x}{\sqrt{{\cos }^{2}x-{\sin }^{2}x}} \end{gathered}$$