Question

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

Answer 1

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

We have $m^2-n^2=(m+n)(m-n)$

$=(\tan \theta +\sin \theta +\tan \theta -\sin \theta )(\tan \theta +\sin \theta -\tan \theta +\sin \theta )$

$=\left(2\tan \theta \right)\left(2\sin \theta \right)$

$=4\tan \theta \sin \theta$ .....$\left(1\right)$

$mn=(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )$

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

$={\sin }^2\theta \left(\frac{1-\cos 2\theta }{{\cos }^2\theta }\right)$

$={\sin }^2\theta \left(\frac{\sin 2\theta }{{\cos }^2\theta }\right)$

$={\sin }^2\theta {\tan }^2\theta$

$\sqrt{mn}=\sqrt{{\sin }^2\theta {\tan }^2\theta }$

$=\sin \theta \tan \theta$

$4\sqrt{mn}=\sin \theta \tan \theta$ .....$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have

$m^2-n^2=4\sqrt{mn}$