Question

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

Answer 1

Given, $m=\tan \theta +\sin \theta ,n=\tan \theta -\sin \theta$

We need to show $m^2-n^2=4\sqrt{mn}$

Taking L.H.S.,

$m^2-n^2=(\tan \theta +\sin \theta {)}^2-(\tan \theta -\sin \theta {)}^2$

$={\tan }^2\theta +{\sin }^2\theta +2\tan \theta .\sin \theta -{\tan }^2\theta -{\sin }^2\theta +2\tan \theta .\sin \theta$

$=4\tan \theta .\sin \theta$

Now, taking R.H.S.,

$4\sqrt{mn}=4\sqrt{(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )}$

$=4\sqrt{{\tan }^2\theta -{\sin }^2\theta }$

$=4\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta }$

$=4\sqrt{\frac{{\sin }^2\theta (1-{\cos }^2\theta )}{{\cos }^2\theta }}$

$=4\sqrt{{\sin }^2\theta .{\tan }^2\theta }$

$=4\tan \theta .\sin \theta$

Therefore, L.H.S. $=$ R.H.S.