Question

If $\cos \theta >0,\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, then show that $m^2-n^2=4\sqrt{mn}$.

Answer 1

$\tan \theta +\sin \theta =m\&\tan \theta -\sin \theta =n$

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

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

$={\sin }^2\theta ({\sec }^2\theta -1)$

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

$\Rightarrow m^2-n^2$

$=(m+n)(m-n)$

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

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

$=4\sqrt{mn}$

Hence, proved.