Question

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

Answer 1

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