Question

If $m=\left(\cos \theta -\sin \theta \right)\mathrm{and}n=\left(\cos \theta +\sin \theta \right)$, then show that $\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\frac{2}{\sqrt{1-{\tan }^2\theta }}$.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}} \\ =\frac{\sqrt{m}}{\sqrt{n}}+\frac{\sqrt{n}}{\sqrt{m}} \\ =\frac{m+n}{\sqrt{mn}} \\ =\frac{\left(\cos \theta -\sin \theta \right)+\left(\cos \theta +\sin \theta \right)}{\sqrt{\left(\cos \theta -\sin \theta \right)\left(\cos \theta +\sin \theta \right)}} \end{gathered}$$
$$\begin{gathered} =\frac{2\cos \theta }{\sqrt{{\cos }^2\theta -{\sin }^2\theta }} \\ =\frac{\left({\displaystyle \frac{2\cos \theta }{\cos \theta }}\right)}{\left({\displaystyle \frac{\sqrt{{\cos }^2\theta -{\sin }^2\theta }}{\cos \theta }}\right)} \\ =\frac{2}{\sqrt{{\displaystyle \frac{{\cos }^2\theta }{{\cos }^2\theta }-\frac{{\sin }^2\theta }{{\cos }^2\theta }}}} \\ =\frac{2}{\sqrt{1-{\tan }^2\theta }} \\ =\mathrm{RHS} \end{gathered}$$