Question

If 2 $\tan A=3\tan B$, Prove that
$\tan (A-B)=\frac{\sin 2B}{5-cos2B}$

Answer 1

$2\tan A=3\tan B$

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$=\frac{\frac{3\tan B}{2}-\tan B}{1+\frac{3}{2}{\tan }^{2}B}=\frac{\tan B}{2\left(\frac{2+3{\tan }^{2}B}{x}\right)}$

.$=\frac{\tan B}{2+3{\tan }^{2}B}=\frac{2\sin B\cos B}{2{\cos }^{2}B}\frac{1}{2+\frac{3{\sin }^{2}B}{{\cos }^{2}B}}$

$=\frac{\sin 2B}{4{\cos }^{2}B+6{\sin }^{2}B}$

$=\frac{\sin 2B}{4{\cos }^{3}B+4{\sin }^{2}B+2{\sin }^{2}B}=\frac{\sin 2B}{4+2{\sin }^{2}B}$

$=\frac{\sin 2B}{5-(1-2{\sin }^{2}B)}=\frac{\sin 2B}{5-\cos 2B}$