三角函数问题1
已知mcos a+sin a=1,ncos a -sin a=1,则m*n=_________
化简(1+cot a-csc a )*(1+tan a+sec a )
过程谢谢
参考答案:mcos a+sin a=1……(1)
ncos a-sin a=1……(2)
(1)+(2)得(m+n)cos a=2 ……(3)
(1)-(2)得(m-n)cos a=-2sin a……(4)
(1)^2-(2)^2得
((m+n)^2-(m-n)^2)cos a^2=4-4sin a^2
4mncos a^2=4cos a^2
mn=1
(1+cot a-csc a )*(1+tan a+sec a )
=(sin a/sin a+cos a/sin a-1/sin a)*(cos a/cos a+sin a/cos a+1/cos a)
=(sin a+cos a-1)/sin a*(cos a+sin a-1)/cos a
=[(sin a+cos a)^2-1]/(sin a*cos a)
=[sin a^2+2sin a*cos a+cos a^2-1]/(sin a*cos a)
=[2sin a*cos a+1-1]/(sin a*cos a)
=(2sin a*cos a)/(sin a*cos a)
=2