求证明一道数学题证明：3+tan1tan2+tan2tan3=tan3/tan1 注：都是角度制的,

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3+tan1tan2+tan2tan3
=1+tan1tan2+1+tan2tan3+1
= (tan2-tan1)/tan(2-1)+ (tan3-tan2)/tan(3-2)+1
=((tan2-tan1)+ (tan3-tan2))/tan1+1
=(tan3-tan1)/tan1+1
=tan3 /tan1-tan1 /tan1+1
=tan3/tan1
======以下答案可供参考======
供参考答案1:
设 tan1=t
则由二倍、三倍角公式
tan2=2t/(1-t^2) tan3=(3t-t^3)/(1-3t^2)
3+tan1tan2+tan2tan3
=3+2t^2/(1-t^2)+2t(3t-t^3)/(1-t^2)(1-3t^2)
=(t^4-4t^2+3)/(t^2-1)(1-3t^2)
=(3-t^2)/(1-3t^2)
=tan3/tan1
证毕