We have:
3 tan (x – 15°) = tan (x + 15°)
⇒ \(\frac{tan(x+15°)}{tan(x-15°)}=3\)
⇒ \(\frac{sin(x+15°)cos(x-15°)}{cos(x+15°)sin(x-15°)}=\frac{3}{1}\)
Applying componendo and dividendo rule, we get
\(\frac{sin(x+15°)cos(x-15°)+cos(x+15°)sin(x-15°)}{sin(x+15°)cos(x-15°)-cos(x+15°)sin(x-15°)}=\frac{3+1}{3-1}\)
\(\frac{sin[(x+15°)+(x-15°)]}{sin[(x+15°)-(x-15°)]}=2\)
\(\frac{sin2x}{sin30°}=2\)
\(\frac{sin2x}{\frac{1}{2}}=2\)
\(sin2x=1\)
\(sin2x=sin\frac{\pi}{2}\)
\(2x=n\pi+(-1)^n\frac{\pi}{2},n∈z\)
\(x=\frac{n\pi}{2}+(-1)^n\frac{\pi}{4},n∈z\)
