Question

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

• A. $\pm \sqrt{mn}$
• B. $\pm 2\sqrt{mn}$
• C. $\pm 4\sqrt{mn}$
• D. $\pm \frac{1}{2}\sqrt{mn}$

Answer 1

The correct option is C $\pm 4\sqrt{mn}$

Given

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

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

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

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

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

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

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

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

from $\left(1\right)$ & $\left(2\right)$, we get

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