Question

Evaluate $\tan {20}^{°}\tan {40}^{°}\tan {60}^{°}\tan {80}^{°}$

• A. $1$
• B. $2$
• C. $3$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is C

$3$

Explanation for correct option :

Evaluating the given trigonometric expression $\tan 20°\tan 40°\tan 60°\tan 80°$

Step-1 : Simplifying the above expression $\tan 20°\tan 40°\tan 60°\tan 80°$

$\tan 20°\times \tan 40°\times \tan 60°\times \tan 80°=\frac{\sin 20°}{\cos 20°}\times \frac{\sin 40°}{\cos 40°}\times \frac{\sin 80°}{\cos 80°}\times \tan 60°\left[\because \tan \theta =\frac{\sin \theta }{\cos \theta }\right]$

Step-2 : Evaluating $\sin 20°\sin 40°\sin 80°$

Formula to be used: We know that $2\sin A\sin B=\cos \left(A-B\right)-\cos \left(A+B\right)$ and $2\cos A\sin B=\sin \left(A+B\right)-\sin \left(A-B\right)$

$\sin 20°\sin 40°\sin 80°=\frac{1}{2}\left(2\sin 20°\sin 40°\right)\sin 80°$.

$$\begin{gathered} =\frac{1}{2}\left(\cos \left(20°-40°\right)-\cos \left(20°+40°\right)\right)\sin 80° \\ =\frac{1}{2}\left(\cos \left(20°\right)-\cos \left(60°\right)\right)\sin 80°\left[\because \cos \left(-\theta \right)=\cos \theta \right] \\ =\frac{1}{4}\left(2\cos 20°\sin 80°\right)-\frac{1}{2}\times \frac{1}{2}\sin 80°\left[\because \cos 60°=\frac{1}{2}\right] \\ =\frac{1}{4}\left(\sin \left(20°+80°\right)-\sin \left(20°-80°\right)\right)-\frac{1}{4}\sin 80° \\ =\frac{1}{4}\left(\sin 100°+\sin 60°-\sin 80°\right)\left[\because \sin \left(-\theta \right)=-\sin \theta \right] \\ =\frac{1}{4}\left(\sin 100°-80°+\frac{\sqrt{3}}{2}\right) \\ =\frac{1}{4}\left(2\cos \frac{100°+80°}{2}\sin \frac{100°-80°}{2}+\frac{\sqrt{3}}{2}\right) \\ =\frac{1}{4}\left(2\cos 90°\sin 10°+\frac{\sqrt{3}}{2}\right)\left[\because \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\right] \\ =\frac{\sqrt{3}}{8}\left[\because \cos 90°=0\right] \end{gathered}$$

Step-3 : Evaluating $\cos 20°\cos 40°\cos 80°$

Formula to be used : We know that $2\cos A\cos B=\cos \left(A+B\right)+\cos \left(A-B\right)$

$\cos 20°\cos 40°\cos 80°=\frac{1}{2}\left(2\cos 20°\cos 40°\right)\cos 80°$

$$\begin{gathered} =\frac{1}{2}\left(\cos \left(20°+40°\right)+\cos \left(20°-40°\right)\right)\cos 80° \\ =\frac{1}{2}\left(\cos 60°+\cos 20°\right)\cos 80°\left[\because \cos \left(-\theta \right)=\cos \theta \right] \\ =\frac{1}{2}\times \frac{1}{2}\cos 80°+\frac{1}{4}\left(2\cos 20°\cos 80°\right)\left[\because \cos 60°=\frac{1}{2}\right] \\ =\frac{1}{4}\cos 80°+\frac{1}{4}\left(\cos \left(20°+80°\right)+\cos \left(20°-80°\right)\right) \\ =\frac{1}{4}\cos 80°+\frac{1}{4}\left(\cos 100°+\cos 60°\right)\left[\because \cos \left(-\theta \right)=\cos \theta \right] \\ =\frac{1}{4}\left(\cos 100°+\cos 80°+\frac{1}{2}\right)\left[\because \cos 60°=\frac{1}{2}\right] \\ =\frac{1}{4}\left(\cos \frac{100°+80°}{2}\cos \frac{100°-80°}{2}+\frac{1}{2}\right) \\ =\frac{1}{4}\left(\cos 90°\cos 10°+\frac{1}{2}\right) \\ =\frac{1}{8}\left[\because \cos 90°=0\right] \end{gathered}$$

Step-4 : Calculating the value of the given expression $\tan {20}^{°}\tan {40}^{°}\tan {60}^{°}\tan {80}^{°}$

Using the values obtained in Step-2 and Step-3, we get :

$\tan 20°\times \tan 40°\times \tan 60°\times \tan 80°=\frac{\sin 20°}{\cos 20°}\times \frac{\sin 40°}{\cos 40°}\times \frac{\sin 80°}{\cos 80°}\times \tan 60°$

$$\begin{gathered} =\frac{{\displaystyle \frac{\sqrt{3}}{8}}}{{\displaystyle \frac{1}{8}}}\sqrt{3}\left[\because \tan 60°=\sqrt{3}\right] \\ =3 \end{gathered}$$

Hence, the correct answer is option (C).