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