求证:[二次根号(x^2+4)][sinx+cosx]/(x^2+5)<1
求证sinx-cosx=根号2sin(x-π/4)
求证 sin^2x/(sinx-cosx)-(sinx+cosx)/tan^2 x-1=sinx+cosx
求证!(1 + cosx )/sinx = cot(x/2)
已知x∈(0,π/2),求证:sinx+cosx>1
如题,求证:sin2x/ [(sinx+cosx-1)(sinx+1-cosx)] =cot(x/2)
求证:tan(x/2)=(1-cosx+sinx)/(1+cosx+sinx)
求证:(1+sinx-cosx)/(1+sinx+cosx)=tan(x/2)
求证:tan(x/2)= sinx/(1+cosx)=(1-cosx)/sinx
f(x)=(根号2sinx+cosx)/sinx+根号(1-sinx)(0
(sin^x/sinx-cosx)-sinx+cosx/tan^2x-1
求导 1/根号x(sinx/2-cosx/2)
求证:tan(x-π/4)=(sinx-cosx)/(sinx+cosx)