证明：1+3sin2αsec4α=sec6α-tan6α（sin、sec、tan后的数字均为平方数,数字后为阿尔法）

Question

证明：1+3sin2αsec4α=sec6α-tan6α（sin、sec、tan后的数字均为平方数,数字后为阿尔法）

Answer 1

证明：令 x =(sin α)^2,
则 (cos α)^2 =1-x,
(sec α)^2 = 1 /(1-x),
(tan α)^2 = x /(1-x).
所以 1 +3 (sin α)^2 (sec α)^4
=1 +3x /(1-x)^2
= (x^2 +x +1) /(1-x)^2.
(sec α)^6 -(tan α)^6
=1 /(1-x)^3 -(x^3) /(1-x)^3
= (1 -x^3) /(1-x)^3
= (1 -x) (1 +x +x^2) /(1-x)^3
= (x^2 +x +1) /(1-x)^2.
所以 1 +3 (sin α)^2 (sec α)^4 =(sec α)^6 -(tan α)^6.
= = = = = = = = =
换元法.
同角三角函数问题,实际上是代数问题.