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

Answer

Given: $3\tan(\text{x}-15^{\circ})=\tan(\text{x}+15^{\circ})$
$\Rightarrow\frac{\tan(\text{x}+15^{\circ})}{\tan(\text{x}-15^{\circ})}=3$
Applying componendo and dividendo, we have
$\frac{\tan(\text{x}+15^{\circ})+\tan(\text{x}-15^{\circ})}{\tan(\text{x}+15)-\tan(\text{x}-15^{\circ})}=\frac{3+1}{3-1}$
$\Rightarrow\frac{\frac{\sin(\text{x}+15^{\circ})}{\cos(\text{x}+15^{\circ})}+\frac{\sin(\text{x}-15^{\circ})}{\cos(\text{x}-15^{\circ})}}{\frac{\sin(\text{x}+15^{\circ})}{\cos(\text{x}+15^{\circ})}-\frac{\sin(\text{x}-15^{\circ})}{\cos(\text{x}-15^{\circ})}}=\frac{4}{2}$
$\Rightarrow\frac{\sin(\text{x}+15^{\circ})\cos(\text{x}-15^{\circ})+\cos(\text{x}+15^{\circ})\sin(\text{x}-15^{\circ})}{\sin(\text{x}+15^{\circ})\cos(\text{x}-15^{\circ})-\cos(\text{x}+15^{\circ})\sin(\text{x}-15^{\circ})}=2$
$\Rightarrow\frac{\sin(\text{x}+15^{\circ}+\text{x}-15^{\circ})}{\sin(\text{x}+15^{\circ}-\text{x}+15^{\circ})}=2$
$\Rightarrow\frac{\sin2\text{x}}{\sin30^{\circ}}=2$
$\Rightarrow\sin2\text{x}=2\times\frac{1}{2}=1$ $\Big(\sin30^{\circ}=\frac{1}{2}\Big)$
$\Rightarrow\sin2\text{x}=\sin90^{\circ}$
$\Rightarrow2\text{x}=90^{\circ}$ $(0<\text{x}<90^{\circ})$
$\Rightarrow\text{x}=45^{\circ}$

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