# If 3tan

Question:

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

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

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