Question

Prove that $\tan 20°\tan 40°\tan 80°=\tan 60°$

Answer 1

Determine the proof $\tan 20°\tan 40°\tan 80°=\tan 60°$.

Use formula:

$$\begin{gathered} 2sinA.sinB=cos(A-B)-cos(A+B) \\ 2cosA.sinB=sin(A+B)-sin(A-B) \\ 2cosA.cosB=cos(A+B)+cos(A-B) \\ \sin C-\sin D=2\cos \frac{C+D}{2}.\sin \frac{C-D}{2} \\ \cos C+\cos D=2\cos \frac{C+D}{2}.\cos \frac{C-D}{2} \end{gathered}$$

Solve the L.H.S part:

$$\begin{gathered} \tan 20°\tan 40°\tan 80°=\frac{\sin 20°}{cos20°}\frac{\sin 40°}{cos40°}\frac{\sin 80°}{cos80°} \\ \Rightarrow =\frac{2\sin 20°\sin 40°\sin 80°}{2cos20°cos40°cos80°} \\ \Rightarrow =\frac{\{\cos (20°-40°)-\cos (20°+40°)\}\sin 80°}{\{cos(20°+40°)+cos(20°-40°)\}cos80°} \\ \Rightarrow =\frac{(\cos 20°-\cos 60°)\sin 80°}{(cos60°+cos20°)cos80°} \\ \Rightarrow =\frac{2\cos 20°\sin 80°-2\left(\frac{1}{2}\right)\sin 80°}{2\left(\frac{1}{2}\right)cos80°+2cos20°cos80°}\because cos60°=\frac{1}{2} \\ \Rightarrow =\frac{\sin (20°+80°)-\sin (20°-80°)-\sin 80°}{cos80°+cos(20°+80°)+cos(20°-80°)} \\ \Rightarrow =\frac{(\sin 100°+\sin 60°-\sin 80°)}{(cos80°+cos100°+cos60°)} \\ \Rightarrow =\frac{2\cos {\displaystyle \frac{(100°+80°)}{2}}\sin {\displaystyle \frac{(100°-80°)}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}}{2cos{\displaystyle \frac{(100°+80°)}{2}}cos{\displaystyle \frac{(100°-80°)}{2}}+\frac{1}{2}} \\ \Rightarrow =\frac{(2\cos 90°\sin 10°+{\displaystyle \frac{\sqrt{3}}{2}})}{(2cos90°cos10°+{\displaystyle \frac{1}{2}})} \\ \Rightarrow =\frac{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}{{\displaystyle \frac{1}{2}}}\because \mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{90}\mathbf{°}=0] \\ \Rightarrow =\sqrt{3} \\ \Rightarrow =tan60° \end{gathered}$$

Hence, the L.H.S = R.H.S.