Question

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, show that $[m^2-n^2{]}^2=16mn$.

Answer 1

${[m^2-n^2]}^2$

$={\left[(m+n)(m-n)\right]}^2$

$={\left[(\tan \theta +\sin \theta +\tan \theta -\sin \theta )(\tan \theta +\sin \theta -\tan \theta +\sin \theta )\right]}^2$

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

$=16{\tan }^2\theta {\sin }^2\theta$ ........$\left(1\right)$

$16mn=16(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )$

$=16({\tan }^2\theta -{\sin }^2\theta )$

$=16{\sin }^2\theta (\frac{1}{{\cos }^2\theta }-1)$

$=16{\sin }^2\theta \left(\frac{1-{\cos }^2\theta }{{\cos }^2\theta }\right)$

$=16{\tan }^2\theta {\sin }^2\theta$ .........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have proved that ${[m^2-n^2]}^2=16mn$