Question

If $\tan x=\frac{b}{a}$, then find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$. [NCERT]

Answer 1

Given: $\tan x=\frac{b}{a}$

$$\begin{gathered} \therefore \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} \\ =\sqrt{\frac{1+{\displaystyle \frac{b}{a}}}{1-{\displaystyle \frac{b}{a}}}}+\sqrt{\frac{1-{\displaystyle \frac{b}{a}}}{1+{\displaystyle \frac{b}{a}}}} \\ =\sqrt{\frac{1+\tan x}{1-\tan x}}+\sqrt{\frac{1-\tan x}{1+\tan x}} \\ =\sqrt{\frac{1+{\displaystyle \frac{\sin x}{\cos x}}}{1-{\displaystyle \frac{\sin x}{\cos x}}}}+\sqrt{\frac{1-{\displaystyle \frac{\sin x}{\cos x}}}{1+{\displaystyle \frac{\sin x}{\cos x}}}} \end{gathered}$$
$$\begin{gathered} =\sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}+\sqrt{\frac{\cos x-\sin x}{\cos x+\sin x}} \\ =\frac{\cos x+\sin x+\cos x-\sin x}{\sqrt{\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)}} \\ =\frac{2\cos x}{\sqrt{{\cos }^{2}x-{\sin }^{2}x}} \\ =\frac{2\cos x}{\sqrt{\cos 2x}} \end{gathered}$$