Question

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

Answer 1

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

$m^2-n^2=(m+n)(m-n)$

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

$=\left(2\tan \theta \right)\left(2\sin \theta \right)$

$=4\tan \theta sin\theta$

$4\sqrt{mn}$

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

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

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

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

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

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

Hence

LHS=RHS (Proved)