Question

If $m=\left(\cos \theta -\sin \theta \right)$ 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

$m=\cos \theta -\sin \theta$ $n=\cos \theta \sin \theta$

$\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\frac{m+n}{\sqrt{mn}}$

$m+n=\cos \theta -\sin \theta +\cos \theta +\sin \theta =2\cos \theta$

$\sqrt{mn}=\sqrt{(\cos \theta -\sin \theta )(\cos \theta +\sin \theta )}$

$=\sqrt{{\cos }^2\theta -{\sin }^2\theta }$

$=\sqrt{\cos 2\theta }$

$\frac{m+n}{\sqrt{mn}}=\frac{2\cos \theta }{\sqrt{\cos 2\theta }}=2\sqrt{\frac{{\cos }^2\theta }{{\cos }^2\theta -{\sin }^2\theta }}$

$=2\sqrt{\frac{1}{\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }}}$

$\sqrt{\frac{m}{n}}+\frac{n}{m}=\frac{2}{\sqrt{1-{\tan }^2\theta }}$