Simplifying the LHS of tan(π3+x)tan(π3−x)=2cos2x+12cos2x−1
tan(π3+x)tan(π3−x)=⎛⎜ ⎜⎝tanπ3+tanx1−tanπ3tanx⎞⎟ ⎟⎠⎛⎜ ⎜⎝tanπ3−tanx1+tanπ3tanx⎞⎟ ⎟⎠
=(tanπ3)2−(tanx)2(1)2−(tanπ3tanx)2
=(√3)2−tan2x1−(√3tanx)2
=3−sin2xcos2x1−3sin2xcos2x
=3cos2x−sin2xcos2x−3sin2x
=3cos2x−(1−cos2x)cos2x−3(1−cos2x)
=3cos2x−1+cos2xcos2x−3+3cos2x
=4cos2x−14cos2x−3
Simplifying the RHS of tan(π3+x)tan(π3−x)=2cos2x+12cos2x−1
2cos2x+12cos2x−1=2(2cos2x−1)+12(2cos2x−1)−1
=4cos2x−2+14cos2x−2−1
=4cos2x−14cos2x−3
Therefore, LHS=RHS.
Hence proved.