If $\tan (\alpha+\beta)=\sqrt{3}, \tan (\alpha-\beta)=1$ where $(\alpha+\beta)$ and $(\alpha-\beta)$ are acute angles, then what is tan $(6 \alpha) ?$

-1

tan60° = √3 & tan 45° = 1

So , ( α + β ) = 60° & ( α - β ) = 45°

on adding ,

2α = 105°

6α = 3 × 105° = 315°

tan6α  = tan315° = tan( 360 - 45°) = -1   ( tan is negative in 4th quadrant )