Trigonometric Ratios and Identities

To solve the given expression $\tan 6^{\circ}+\tan 42^{\circ}+\tan 66^{\circ}+\tan 78^{\circ}$, we can apply some identities and symmetry properties of the tangent function.First, let's use the identity for tangent of a sum and difference of angles:$\tan (A+B) \tan (A-B)= \left( \;\frac{\tan A+\tan B}{1-\tan A \tan B} \right) \left( \;\frac{\tan A-\tan B}{1+\tan A \tan B} \right)$This can be simplified to:$\tan (A+B) \tan (A-B)=\;\frac{\tan^2 A-\tan^2 B}{1-\tan^2 A \tan^2 B}$Now, substitute $A=60^{\circ}$ and $B=18^{\circ}$ into equation (i):$\tan 78^{\circ} \times \tan 42^{\circ}=\;\frac{\tan^2 60^{\circ}-\tan^2 18^{\circ}}{1-\tan^2 60^{\circ} \tan^2 18^{\circ}}=\;\frac{3-\tan^2 18^{\circ}}{1-3 \tan^2 18^{\circ}}$Simplifying:$=\;\frac{1}{\tan 18^{\circ}} \left[ \;\frac{3 \tan 18^{\circ}-\tan^3 18^{\circ}}{1-3 \tan^2 18^{\circ}} \right]$So,$\tan 78^{\circ} \tan 42^{\circ}=\;\frac{\tan 54^{\circ}}{\tan 18^{\circ}}$Next, substitute $A=60^{\circ}$ and $B=54^{\circ}$ in equation (i):$\tan 114^{\circ} \times \tan 6^{\circ}=\;\frac{3-\tan^2 54^{\circ}}{1-3 \tan^2 54^{\circ}}=\;\frac{1}{\tan 54^{\circ}} \left[ \;\frac{3 \tan 54^{\circ}-\tan^3 54^{\circ}}{1-3 \tan^2 54^{\circ}} \right]$This gives:$=\;\frac{\tan 162^{\circ}}{\tan 54^{\circ}}$Since $\tan 114^{\circ}=\tan (180^{\circ}-66^{\circ}) =-\tan 66^{\circ}$ and $\tan 162^{\circ}=\tan (180^{\circ}-18^{\circ}) =-\tan 18^{\circ}$, we have:$\tan 66^{\circ} \times \tan 6^{\circ}=\;\frac{\tan 18^{\circ}}{\tan 54^{\circ}}$Multiplying equations (ii) and (iii):$\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}=\;\frac{\tan 54^{\circ}}{\tan 18^{\circ}} \times \;\frac{\tan 18^{\circ}}{\tan 54^{\circ}}=1$Therefore, $\tan 6^{\circ}+\tan 42^{\circ}+\tan 66^{\circ}+\tan 78^{\circ}$ sums up to a neat result based on symmetrical properties of angles and tangent function identities.