Question

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, then show that $m^2-n^2=4\sqrt{mn}$ .

Answer 1

$\tan \theta +\sin \theta =m\tan \theta -\sin \theta =nFromtheaboveequations,weget\tan \theta =\frac{m+n}{2},\sin \theta =\frac{m-n}{2}$

$\tan \theta +\sin \theta =m\Rightarrow \tan \theta (1+\cos \theta )=m\left(1\right)\tan \theta -\sin \theta =n\Rightarrow \tan \theta (1-\cos \theta )=n\left(2\right)\frac{1+\cos \theta }{1-\cos \theta }=\frac{m}{n}\Rightarrow n+n\cos \theta =m-m\cos \theta \Rightarrow \cos \theta =\frac{m-n}{m+n}\Rightarrow {\cos }^2\theta =1-{\sin }^2\theta ={\left(\frac{m-n}{m+n}\right)}^2\Rightarrow 1-{\left(\frac{m-n}{2}\right)}^2={\left(\frac{m-n}{m+n}\right)}^2\Rightarrow {(m+n)}^2-{\left(\frac{m^2-n^2}{2}\right)}^2={(m-n)}^2\Rightarrow 4mn=\frac{{(m^2-n^2)}^2}{4}\Rightarrow 16mn={(m^2-n^2)}^2\Rightarrow (m^2-n^2)=\sqrt{16mn}=4\sqrt{mn}$