Question

If $2\tan \alpha =3\tan \beta ,\mathrm{then}\tan \left(\alpha -\beta \right)=$

(a) $\frac{\sin 2\beta }{5-\cos 2\beta }$

(b) $\frac{\cos 2\beta }{5-\cos 2\beta }$

(c) $\frac{\sin 2\beta }{5+\cos 2\beta }$

(d) none of these

Answer 1

(a) $\frac{\sin 2\beta }{5-\cos 2\beta }$

$$\begin{gathered} \mathrm{Given}: \\ 2\tan \alpha =3\tan \beta \\ \mathrm{Now}, \\ \tan \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\frac{\mathrm{tan\alpha }-\mathrm{tan\beta }}{1+\mathrm{tan\alpha tan\beta }} \\ =\frac{{\displaystyle \frac{3}{2}}\mathrm{tan\beta }-\mathrm{tan\beta }}{1+\left({\displaystyle \frac{3}{2}}\mathrm{tan\beta }\right)\mathrm{tan\beta }} \\ =\frac{{\displaystyle 3}\mathrm{tan\beta }-2\mathrm{tan\beta }}{2+3{\tan }^2\mathrm{\beta }} \\ =\frac{\mathrm{tan\beta }}{2+3{\tan }^2\mathrm{\beta }} \\ =\frac{{\displaystyle \frac{\mathrm{sin\beta }}{\mathrm{cos\beta }}}}{2+3{\displaystyle \frac{{\sin }^2\mathrm{\beta }}{{\cos }^2\mathrm{\beta }}}} \\ =\frac{\mathrm{sin\beta cos\beta }}{2{\cos }^2\mathrm{\beta }+3{\sin }^2\mathrm{\beta }} \\ =\frac{\mathrm{sin\beta cos\beta }}{2+{\sin }^2\mathrm{\beta }} \\ =\frac{2\mathrm{sin\beta cos\beta }}{4+2{\sin }^2\mathrm{\beta }} \\ =\frac{\sin 2\mathrm{\beta }}{4+1-\cos 2\mathrm{\beta }} \\ \therefore \tan \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\frac{\sin 2\mathrm{\beta }}{5-\cos 2\mathrm{\beta }} \end{gathered}$$