Question

Solve $3\tan (\theta -{15}^o)=\tan (\theta +{15}^o)$.

Answer 1

$\frac{\tan A}{\tan B}=\frac{3}{1}$, where A$=\theta +{15}^o,B=\theta -{15}^o$
Apply componeudo and dividend
$\therefore \frac{\tan A+\tan B}{\tan A-\tan B}=\frac{3+1}{3-1}$ or $\frac{\sin (A+B)}{\sin (A-B)}=2$
or $\sin 2\theta =2\sin {30}^o=2\cdot \frac{1}{2}=1=\sin \frac{\pi }{2}$
$\therefore 2\theta =2n\pi +\frac{\pi }{2}$ or $\theta =n\pi +\frac{\pi }{4}$.