请教一道高一数学三角函数证明题

左式上下同乘(cosx+sinx+1)，则

(cosx+sinx+1)^2
左式=-------------------
(cosx+1)^2-(sinx)^2

(cosx)^2+(sinx)^2+1+2cosx+2sinx+2sinxcosx
=-----------------------------------------
(cosx)^2+1+2cosx-(sinx)^2

2[1+cosx+sinx+sinxcosx]
=-----------------------
2[(cosx)^2+cosx]

(1+cosx)(1+sinx)
=----------------
cosx(cos+1)

1+sinx
=------
cosx

=右式

左边=(cosx+sinx+1)/(cosx-sinx+1)=[2cos²（x/2)+2sin(x/2)cos(x/2)]/[2cos²（x/2)-2sin(x/2)cos(x/2)]=[cos（x/2)+sin(x/2)]／[cos（x/2)-sin(x/2)]
右边=（1+sinx）/cosx=[sin(x/2)+cos(x/2)]²/[cos²（x/2)-sin²(x/2)]=[cos（x/2)+sin(x/2)]／[cos（x/2)-sin(x/2)]
∴左边=右边