1,Sin(a+π/4)/(Sin2a+Cos2a+1)= [2分之根号2*（sina+cosa）]/（sin2a+2cos平方a）=[2分之根号2*（sina+cosa）]/（2sinacosa+2cos平方 a）=[2分之根号2*（sina+cosa）]/[2cosa*(sina+cosa)]=2分之根号2/(cosa*2)
由已知条件得cosa=-1/4带入得 ：负根号2
2,=tan²x+1/tan²x
=sin²x/cos²x+cos²x/sin²x
=[(sinx)^4+(cosx)^4]/(sinxcosx)²
={[(sinx)²+(cosx)²]²-2(sinxcosx)²}/(sinxcosx)²
=[1-2(sinxcosx)²]/(sinxcosx)²
=4[1-1/2(sin2x)²]/(sin2x)²
=2[4-2(sin2x)²]/(1-cos4x)
=2(3+cos4x)/(1-cos4x)
3,tanb=2tan(b/2)/(1-tan平方b/2)=4/3