Question

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, then the value of $m^2-n^2$ is equal to -

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

Answer 1

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

$\tan \theta +\sin \theta =m$

$\Rightarrow \tan \theta -\sin \theta =n$

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

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

$=2\tan \theta .2\sin \theta$

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

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

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

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

$={\sin }^2\theta [\frac{1}{{\cos }^2\theta }-1]$

$={\sin }^2\theta \left[\frac{1-{\cos }^2\theta }{{\cos }^2\theta }\right]$

$={\sin }^2\theta .\frac{{\sin }^2\theta }{{\cos }^2\theta }$

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

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

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

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

Hence, the answer is $4\sqrt{mn}.$