If 3tan
Question:

If $3 \tan \left(x-15^{\circ}\right)=\tan \left(x+15^{\circ}\right), 0<x<90^{\circ}$, find $\theta$.

Solution:

Given: $3 \tan \left(x-15^{\circ}\right)=\tan \left(x+15^{\circ}\right)$

$\Rightarrow \frac{\tan \left(x+15^{\circ}\right)}{\tan \left(x-15^{\circ}\right)}=3$

Applying componendo and dividendo, we have

$\frac{\tan \left(x+15^{\circ}\right)+\tan \left(x-15^{\circ}\right)}{\tan \left(x+15^{\circ}\right)-\tan \left(x-15^{\circ}\right)}=\frac{3+1}{3-1}$

$\Rightarrow \frac{\frac{\sin \left(x+15^{\circ}\right)}{\cos \left(x+15^{\circ}\right)}+\frac{\sin \left(x-15^{\circ}\right)}{\cos \left(x-15^{\circ}\right)}}{\frac{\sin \left(x+15^{\circ}\right)}{\cos \left(x+15^{\circ}\right)}-\frac{\sin \left(x-15^{\circ}\right)}{\cos \left(x-15^{\circ}\right)}}=\frac{4}{2}$

$\Rightarrow \frac{\sin \left(x+15^{\circ}\right) \cos \left(x-15^{\circ}\right)+\cos \left(x+15^{\circ}\right) \sin \left(x-15^{\circ}\right)}{\sin \left(x+15^{\circ}\right) \cos \left(x-15^{\circ}\right)-\cos \left(x+15^{\circ}\right) \sin \left(x-15^{\circ}\right)}=2$

$\Rightarrow \frac{\sin \left(x+15^{\circ}+x-15^{\circ}\right)}{\sin \left(x+15^{\circ}-x+15^{\circ}\right)}=2$

$\Rightarrow \frac{\sin 2 x}{\sin 30^{\circ}}=2$

$\Rightarrow \sin 2 x=2 \times \frac{1}{2}=1$   $\left(\sin 30^{\circ}=\frac{1}{2}\right)$

$\Rightarrow \sin 2 x=\sin 90^{\circ}$

$\Rightarrow 2 x=90^{\circ}$      $\left(0<x<90^{\circ}\right)$

$\Rightarrow x=45^{\circ}$