高一数学已知(2+cot^2Q)/(1+sinQ)=1,那么(1
已知(2+cot^2Q)/(1+sinQ)=1,那么(1+sinQ)(2+ cQ)=( )
已知(2+cot^2Q)/(1+sinQ)=1,则: [2+(cos^Q/sin^Q)]/(1+sinQ)=1 ===> (2sin^Q+cos^Q)/(sin^Q+sin^3Q)=1 ===> (1+sin^Q)/(sin^Q+sin^3Q)=1 ===> 1+sin^Q=sin^Q+sin^3Q ===> sin^3Q=1 ===> sinQ=1,且cosQ=0.则: ===> (1+sinQ)(2+coaQ)=(1+1)(2+0)=4