Question

If $\tan (\alpha -\beta )=\frac{\sin 2\beta }{5-\cos 2\beta }$, find the value of $\frac{\tan \alpha }{\tan \beta }$.

Answer 1

Given that-

$2\tan \alpha =3\tan \beta$

$\tan \alpha =\frac{3}{2}$

$tan(\alpha -\beta )=\frac{tan\alpha -tan\beta }{1+tan\alpha tan\beta }$

$=\frac{\frac{3}{2}tan\beta -tan\beta }{1+\frac{3}{2}ta{n}^2\beta }$

$=\frac{\tan \beta }{1+\frac{3}{2}{\tan }^2\beta }$

$=\frac{\frac{\sin \beta }{\cos \beta }}{1+\frac{3}{2}\frac{{\sin }^2\beta }{{\cos }^2\beta }}$

$=\frac{\sin \beta \cos \beta }{2{\cos }^2\beta +3{\sin }^2\beta }$

$=\frac{2\sin \beta \cos \beta }{2((1+\cos 2\beta )+\frac{3}{2}(1-\cos 2\beta ))}$

$=\frac{\sin 2\beta }{5-\cos 2\beta }$

Hence, proved .