(1)求A
[(tan(派/4-x)(sin^2[(派/4+x)]
=cot[派/4+x](sin^2[(派/4+x)]
=2sin(π/4+x)cos(π/4+x)
=sin2(π/4+x)
=sin(π/2+2x)
=cos2x
A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2[(派/4)+x)] ]
=cos2x/cos2x
=1
(2)tanB=1,所以B=kπ+π/4
4B=4kπ+π,k整数.
已知A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2(派/4)+x)]
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