Question

Let $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$ then show that ${\tan }^2\theta -{\sin }^2\theta =mn$.

Answer 1

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

Now,

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

or, ${\tan }^2\theta -{\sin }^2\theta =mn$

or, $(\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta )=mn$

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

or, $\frac{{\sin }^2\theta }{{\cos }^2\theta }\cdot {\sin }^2\theta =mn$ $[\because 1-{\cos }^2\theta ={\sin }^2\theta ]$

or, ${\tan }^2\theta \cdot {\sin }^2\theta =mn$.