若实数x,y满足 x+y/1-xy=根号2,求 |1-xy|/（根号下1+x²）×（根号下1+y²）

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若实数x,y满足 x+y/1-xy=根号2,求 |1-xy|/（根号下1+x²）×（根号下1+y²）思路是用三角函数解

x = tan(a),y = tan(b)
那么sqrt(2) = (x+y)/(1-xy) = (tana + tanb) / (1 - tana*tanb) = tan(a+b)
|1-xy|/...= |1-tana*tanb| / (|seca| * |secb|) = |1-tana*tanb| * |cosa| * |cosb|
= |cosa * cosb - sina * sinb| = |cos(a+b)| = 1/sqrt(1 + tan^2(a+b)) = 1/sqrt(3)

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