Prove that

Solution:

To Prove: $\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}=1$

Taking LHS,

$=\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}$

Multiply and divide by tan 54° tan 18

$=\frac{\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}}{\tan 54^{\circ} \tan 18^{\circ}} \times \tan 54^{\circ} \tan 18^{\circ}$

$=\frac{\left(\tan 6^{\circ} \tan 54^{\circ} \tan 66^{\circ}\right)\left(\tan 18^{\circ} \tan 42^{\circ} \tan 72^{\circ}\right)}{\tan 54^{\circ} \tan 18^{\circ}} \ldots$ (i)

We know that

$\tan x \tan \left(60^{\circ}-x\right) \tan \left(60^{\circ}+x\right)=\tan 3 x$

So, eq. (i) becomes

$=\frac{\left[\tan 3\left(6^{\circ}\right)\right]\left[\tan 3\left(18^{\circ}\right)\right]}{\tan 54^{\circ} \tan 18^{\circ}}$

$=\frac{\tan 18^{\circ} \tan 54^{\circ}}{\tan 54^{\circ} \tan 18^{\circ}}$

$=1$

$=\mathrm{RHS}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence Proved