Question

If $3\tan \left(\theta -15°\right)=\tan \left(\theta +15°\right)$, $0<\theta <90°$, find $\theta$.

Answer 1

Given: $3\tan \left(\theta -15°\right)=\tan \left(\theta +15°\right)$

$\Rightarrow \frac{\tan \left(\theta +15°\right)}{\tan \left(\theta -15°\right)}=3$

Applying componendo and dividendo, we have

$$\begin{gathered} \frac{\tan \left(\theta +15°\right)+\tan \left(\theta -15°\right)}{\tan \left(\theta +15°\right)-\tan \left(\theta -15°\right)}=\frac{3+1}{3-1} \\ \Rightarrow \frac{{\displaystyle \frac{\sin \left(\theta +15°\right)}{\cos \left(\theta +15°\right)}}+{\displaystyle \frac{\sin \left(\theta -15°\right)}{\cos \left(\theta -15°\right)}}}{{\displaystyle \frac{\sin \left(\theta +15°\right)}{\cos \left(\theta +15°\right)}}-{\displaystyle \frac{\sin \left(\theta -15°\right)}{\cos \left(\theta -15°\right)}}}=\frac{4}{2} \\ \Rightarrow \frac{\sin \left(\theta +15°\right)\cos \left(\theta -15°\right)+\cos \left(\theta +15°\right)\sin \left(\theta -15°\right)}{\sin \left(\theta +15°\right)\cos \left(\theta -15°\right)-\cos \left(\theta +15°\right)\sin \left(\theta -15°\right)}=2 \\ \Rightarrow \frac{\sin \left(\theta +15°+\theta -15°\right)}{\sin \left(\theta +15°-\theta +15°\right)}=2 \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{\sin 2\theta }{\sin 30°}=2 \\ \Rightarrow \sin 2\theta =2\times \frac{1}{2}=1\left(\sin 30°=\frac{1}{2}\right) \\ \Rightarrow \sin 2\theta =\sin 90° \\ \Rightarrow 2\theta =90°\left(0<\theta <90°\right) \\ \Rightarrow \theta =45° \end{gathered}$$