Solution: The original formula = tan (20+40) (1-tan 20 tan 40)+√ 3 tan 20 tan 40.

=√3( 1-tan 20 tan 40)+√3 tan 20 tan 40

=√3

2.tan(α+β)=tan3π/4=- 1

tan(α+β)=(tanα+tanβ)/( 1-tanαtanβ)=- 1

tanα+tanβ=- 1+tanαtanβ

tanα+tanβ-tanαtanβ=- 1

( 1-tanα)( 1-tanβ)

=tanαtanβ-(tanα+tanβ)+ 1

= 1+ 1

=2

3. The original formula = [tan 60 (1-tan 20 * tan 40 *)+tan (-60)]/tan 20 * tan 40.

=[tan 60(-tan 20 * tan 40)]/tan 20 * tan 40

=-tan60

=-√3