Question

If $tan\theta +sin\theta =m$ and $tan\theta -sin\theta =n$ then $m^2-n^2=$

• A. $\sqrt{mn}$
• B. mn
• C. 2mn
• D. $4\sqrt{mn}$

Answer 1

The correct option is D $4\sqrt{mn}$

$m^2-n^2$

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

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

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

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

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

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

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

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