If $3tan(\theta -15°)=tan(\theta +15°),0<\theta <\pi \text{ then }\theta \text{=}$

$3\tan (\theta -15°)=\tan (\theta +15°)$

$\begin{array}{l}\frac{3}{1}=\frac{\tan (\theta +15°)}{\tan (\theta -15°)}\\ \frac{4}{2}=\frac{\tan (\theta +15°)+\tan (\theta -15°)}{\tan (\theta +15°)-\tan (\theta -15°)}\\ 2=\frac{\frac{\sin (\theta +15)}{\cos (\theta +15)}+\frac{\sin (\theta -15)}{\cos (\theta -15)}}{\frac{\sin (\theta +15)}{\cos (\theta +15)}-\frac{\sin (\theta -15)}{\cos (\theta -15)}}\\ 2=\frac{\sin (\theta +15°+\theta -15°)}{\sin (\theta +15°-\theta +15°)}\\ 2=\frac{\sin 2\theta }{\sin 30°}\\ \sin 2\theta =1\\ 2\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{4}\end{array}$