Question

If $2\tan \alpha =3\tan \beta ,$ prove that $\tan \left(\alpha -\beta \right)=\frac{\sin 2\beta }{5-\cos 2\beta }$

Answer 1

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

$$\begin{gathered} \mathrm{LHS}=\frac{\mathrm{tan\alpha }-\mathrm{tan\beta }}{1+\mathrm{tan\alpha }\times \mathrm{tan\beta }} \\ =\frac{{\displaystyle \frac{3}{2}}\times \mathrm{tan\beta }-\mathrm{tan\beta }}{1+{\displaystyle \frac{3}{2}}{\tan }^2\mathrm{\beta }}\left(\because 2\mathrm{tan\alpha }=3\mathrm{tan\beta }\right) \\ =\frac{{\displaystyle \frac{1}{2}}\times \mathrm{tan\beta }}{1+{\displaystyle \frac{3}{2}}{\tan }^2\mathrm{\beta }}=\frac{\mathrm{tan\beta }}{2+3{\tan }^2\mathrm{\beta }} \\ =\frac{{\displaystyle \frac{\mathrm{sin\beta }}{\mathrm{cos\beta }}}}{2+3{\displaystyle \frac{{\sin }^2\mathrm{\beta }}{{\cos }^2\mathrm{\beta }}}}=\frac{{\displaystyle \frac{\mathrm{sin\beta }}{\mathrm{cos\beta }}}\times {\cos }^2\mathrm{\beta }}{2{\cos }^2\mathrm{\beta }+3{\sin }^2\mathrm{\beta }} \end{gathered}$$

$$\begin{gathered} =\frac{\mathrm{sin\beta }\times \mathrm{cos\beta }}{2{\cos }^2\mathrm{\beta }+2{\sin }^2\mathrm{\beta }+{\sin }^2\mathrm{\beta }} \\ =\frac{1}{2}\frac{2\mathrm{sin\beta }\times \mathrm{cos\beta }}{2\left({\cos }^2\mathrm{\beta }+{\sin }^2\mathrm{\beta }\right)+{\sin }^2\mathrm{\beta }} \\ =\frac{1}{2}\frac{\sin 2\mathrm{\beta }}{\left(2+{\sin }^2\mathrm{\beta }\right)}=\frac{\sin 2\mathrm{\beta }}{4+2{\sin }^2\mathrm{\beta }} \\ =\frac{\sin 2\mathrm{\beta }}{4+2\left(1-{\cos }^2\mathrm{\beta }\right)}=\frac{\sin 2\mathrm{\beta }}{6-2{\cos }^2\mathrm{\beta }} \\ =\frac{\sin 2\mathrm{\beta }}{6-\left(1+\cos 2\mathrm{\beta }\right)}\left(\because 1+\cos 2\mathrm{\beta }=2{\cos }^2\mathrm{\beta }\right) \\ =\frac{\sin 2\mathrm{\beta }}{5-\cos 2\mathrm{\beta }}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$