Question

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

Answer 1

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