(2)已知sin2Acsc2B+cos2Acos2C=1(cosA≠0),求证:sin2C=tan2Acot2B.
证明:(1)右边=
=
=左边.
(2)∵sin2Acsc2B+cos2A(1-sin2C)=1,∴cos2Asin2C=sin2Acsc2B-(1-cos2A).
∴sin2C=(csc2B-1)=tan2A·cot2B.