Question

If $m\sin \theta =n\cos \theta$ then show that :$\frac{\tan \theta +\cot \theta }{\tan \theta -\cot \theta }=\frac{n\sin \theta +m\cos \theta }{n\sin \theta -m\cos \theta }=\frac{n^2+m^2}{n^2-m^2}$

Answer 1

$m\sin \theta =n\cos \theta \Rightarrow \sin \theta =\frac{n}{m}\cos \theta$

$\Rightarrow \tan \theta =\frac{n}{m}\Rightarrow \cot \theta =\frac{m}{n}$

$\frac{\tan \theta +\cot \theta }{\tan \theta -\cot \theta }=\frac{\frac{n}{m}+\frac{m}{n}}{\frac{n}{m}-\frac{m}{n}}$

$=\frac{\left(\frac{n^2+m^2}{mn}\right)}{\left(\frac{n^2-m^2}{mn}\right)}$

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

$\frac{n\sin \theta +m\cos \theta }{n\sin \theta -m\cos \theta }=\frac{n\left(\frac{n}{m}\right)\cos \theta +m\cos \theta }{[{}^n\left(\frac{n}{m}\right)\cos \theta -m\cos \theta ]}$

$=\frac{\cos \theta [\frac{n^2}{m}+m]}{\cos \theta [\frac{n^2}{m}-m]}$

$=\frac{(n^2+m^2)}{(n^2-m^2)}$

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

$\Rightarrow \frac{\tan \theta +\cot \theta }{\tan \theta -\cot \theta }=\frac{n\sin \theta +m\cos \theta }{m\sin \theta -m\cos \theta }=\frac{n^2+m^2}{n^2-m^2}$