Question

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, then $(m^2-n^2)=4\sqrt{mn}$.

• A. True
• B. False

Answer 1

The correct option is A True

LHS :

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

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

RHS :

$4\sqrt{mn}=4\sqrt{(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )}$

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

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

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

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

Thus, LHS = RHS

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