Prove $\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}=\tan85^{\circ}$ (Method II)

$\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}$

$=\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}\cdot \dfrac{\tan85^{\circ}}{\tan85^{\circ}}$

$=\tan85^{\circ}\cdot \dfrac{\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}}{\tan85^{\circ}}$

$=\tan85^{\circ}\cdot \dfrac{\sin 55^{\circ}\sin 65^{\circ}\sin 75^{\circ}(\cos85^{\circ})}{\cos 55^{\circ}\cos 65^{\circ}\cos 75^{\circ}\sin85^{\circ}}$

$=\tan85^{\circ}\cdot \dfrac{(\sin 55^{\circ}\sin 65^{\circ})(\sin 75^{\circ}\cos85^{\circ})}{(\cos 55^{\circ}\cos 65^{\circ})(\cos 75^{\circ}\sin85^{\circ})}$

$=\tan85^{\circ}\cdot\dfrac{{-\dfrac{1}{2}(\cos 120^{\circ}-\cos 10^{\circ})\frac{1}{2}(\sin160^{\circ}-\sin10^{\circ})}}{{\dfrac{1}{2}(\cos 120^{\circ}+\cos 10^{\circ})\dfrac{1}{2}(\sin160^{\circ}+\sin10^{\circ})}}$

$=-\tan85^{\circ}\cdot \dfrac{(\cos 120^{\circ}-\cos 10^{\circ})(\sin160^{\circ}-\sin10^{\circ})}{(\cos 120^{\circ}+\cos 10^{\circ})(\sin160^{\circ}+\sin10^{\circ})}$

$=-\tan85^{\circ}\cdot \dfrac{-\dfrac{1}{2}\sin160^{\circ}+\frac{1}{2}\sin10^{\circ}-\sin160^{\circ}\cos 10^{\circ}+\sin10^{\circ}\cos10^{\circ}}{-\frac{1}{2}\sin160^{\circ}-\frac{1}{2}\sin10^{\circ}+\sin160^{\circ}\cos 10^{\circ}+\sin10^{\circ}\cos10^{\circ}}$

$=-\tan85^{\circ}\cdot \dfrac{-\frac{1}{2}\sin20^{\circ}+\dfrac{1}{2}\sin10^{\circ}-\sin160^{\circ}\cos 10^{\circ}+\dfrac{1}{2}\sin20^{\circ}}{-\dfrac{1}{2}\sin20^{\circ}-\dfrac{1}{2}\sin10^{\circ}+\sin160^{\circ}\cos 10^{\circ}+\dfrac{1}{2}\sin20^{\circ}}$

$=-\tan85^{\circ}\cdot \dfrac{\dfrac{1}{2}\sin10^{\circ}-\sin160^{\circ}\cos 10^{\circ}}{-\frac{1}{2}\sin10^{\circ}+\sin160^{\circ}\cos 10^{\circ}}$

$=-\tan85^{\circ}\cdot (-1)$

$=\tan85^{\circ}$ (Q.E.D.)