Question

The value of:
$\tan 6^{\circ }\tan {42}^{\circ }\tan {66}^{\circ }\tan {78}^{\circ }$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

$\tan 6°\tan 42°\tan 66°\tan 78°$

$\Rightarrow \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\Rightarrow \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$\therefore \tan (A+B).\tan (A-B)=\frac{{\tan }^{2}A-{\tan }^{2}B}{1-{\tan }^{2}A{\tan }^{2}B}$

Let $A=60°,$ then

$\Rightarrow \tan (60°-B)\tan (60°+B)=\frac{{\left(\tan 60°\right)}^2-{\tan }^{2}B}{1-{\tan }^{2}60°{\tan }^{2}B}$

$=\frac{3-{\tan }^{2}B}{1-3{\tan }^{2}B}$

$=\frac{\tan B}{\tan B}.\frac{3-{\tan }^{2}B}{1-3{\tan }^{2}B}$

$=\frac{1}{\tan B}.\frac{3\tan B-{\tan }^{3}B}{1-3{\tan }^{3}B}$

$=\frac{\tan 2B}{\tan B}$

$\Rightarrow \tan (60°-B)\tan B\tan (60°+B)=\tan 3B$

When $B=18°$

$\Rightarrow \tan 42°\tan 18°\tan 78°=\tan 54°$

$\Rightarrow \tan 42°\tan 78°=\frac{\tan 54°}{\tan 18°}⟶\left(1\right)$

When $B=54°$

$\Rightarrow \tan 6°\tan 54°\tan 114°=\tan 162°$

$\Rightarrow \tan 6°\tan 54°\tan (180°-114°)=\tan (180°-162°)$

$\Rightarrow \tan 6°\tan 66°=\frac{\tan 18°}{\tan 54°}⟶\left(2\right)$

Multiply $\left(1\right)\xi \left(2\right),$ we get

$\Rightarrow \tan 6°\tan 42°\tan 66°\tan 78°=\frac{\tan 54°}{\tan 18°}.\frac{\tan 18°}{\tan 54°}$

$\Rightarrow \tan 6°\tan 42°\tan 66°\tan 78°=1$

Hence, the answer is $1.$