27.(2023春•大兴区期中)已知函数f(x)=excosx−x−1f(x)=e^{x}\cos x-x-1f(x)=excosx−x−1.(Ⅰ)求曲线y=f(x)y=f(x)y=f(x)在点(0(0(0,f(0))f(0))f(0))处的切线方程;(Ⅱ)设g(x)=f′(x)g(x)=f\prime (x)g(x)=f′(x),求证:当x∈[0x\in [0x∈[0,π)\pi )π)时,g(x)⩽0g(x)\leqslant 0g(x)⩽0;(Ⅲ)对任意的m , n∈(0 , π2)m\;,\;n\in (0\;,\;\frac{\pi }{2})m,n∈(0,2π),判断f(m+n)−f(m)f(m+n)-f(m)f(m+n)−f(m)与f(n)f(n)f(n)的大小关系,并证明结论.