（1）在极坐标系中，已知圆ρ=2cosθ与直线3ρcosθ+4ρsinθ+a=0相切，求实数a的值.
（2）对5副不同的手套进行不放回抽取，甲先任取一只，乙再任取一只，然后甲又任取一只，最后乙再任取一只．对于下列事件：①A：甲正好取得两只配对手套；②B：乙正好取得两只配对手套.试判断事件A与B是否独立？并证明你的结论．
（1）
，或
. （2）
≠
， A与B是不独立的．
题目分析：
，圆ρ=2cosθ的普通方程为：
，即
直线3ρcosθ+4ρsinθ+a=0的普通方程为：
， 4分
又圆与直线相切，所以
解得：，或. 7分
（2）解：①P（A）= 
= ；
②
== ． 11分
∵P（AB）= = , =
， 13分
∴≠，故A与B是不独立的． 15分
点评：中档题，本题综合性较强，覆盖面较广。考查知识点注重了基础。其中（1）化为直角坐标方程，利用几何法研究直线与圆相切问题，是常见方法。相互独立事件的概率满足=。
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