Question

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

• A. $1$
• B. $4$
• C. $8$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $1$

$\tan 6^{\circ }\tan {42}^{\circ }\tan {66}^{\circ }\tan {78}^{\circ }=\tan 6^{\circ }\tan \left({66}^{\circ }\right)\tan ({60}^{\circ }-{18}^{\circ })\tan ({60}^{\circ }+{18}^{\circ })$
Multiply and divide by $\tan {18}^{\circ }$, we get
$=\frac{\tan 6^{\circ }\tan \left({66}^{\circ }\right)\tan {18}^{\circ }\tan ({60}^{\circ }-{18}^{\circ })\tan ({60}^{\circ }+{18}^{\circ })}{\tan {18}^{\circ }}=\frac{\tan 6^{\circ }\tan \left({66}^{\circ }\right)\tan (3\times {18}^{\circ })}{\tan {18}^{\circ }}=\frac{\tan 6^{\circ }\tan \left({66}^{\circ }\right)\tan \left({54}^{\circ }\right)}{\tan {18}^{\circ }}=\frac{\tan 6^{\circ }\tan ({60}^{\circ }+6^{\circ })\tan ({60}^{\circ }-6^{\circ })}{\tan {18}^{\circ }}=\frac{\tan 3\times 6^{\circ }}{\tan {18}^{\circ }}=\frac{\tan {18}^{\circ }}{\tan {18}^{\circ }}=1$