化简tan2αtan(π/6-α)+tan2αtan(π/3-α)+tan(π/6-α)tan(π/3-α)=?请写下过程,

题目

化简tan2αtan(π/6-α)+tan2αtan(π/3-α)+tan(π/6-α)tan(π/3-α)=?请写下过程,

答案

知识：tan(a+b)=(tana+tanb)/(1-tanatanb)
原式=tan2a[tan(π/6-a)+tan(π/3-a)]+tan(π/6-a)tan(π/3-a)
=tan2a*tan(π/6-a+π/3-a)*[1-tan(π/6-a)tan(π/3-a)]+tan(π/6-a)tan(π/3-a)
=tan2a*tan(π/2-2a)*[1-tan(π/6-a)tan(π/3-a)]+tan(π/6-a)tan(π/3-a)
=tan2a*cot2a*[1-tan(π/6-a)tan(π/3-a)]+tan(π/6-a)tan(π/3-a)
=1-tan(π/6-a)tan(π/3-a)+tan(π/6-a)tan(π/3-a)
=1