# If 2 tan α=3 tan β,

Question:

If $2 \tan \alpha=3 \tan \beta$, then $\tan (\alpha-\beta)=$

(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

Solution:

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

Given:

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

Now,

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

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

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

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

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

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

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

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

$=\frac{\sin 2 \beta}{4+1-\cos 2 \beta}$

$\therefore \tan (\alpha-\beta)=\frac{\sin 2 \beta}{5-\cos 2 \beta}$