拜托把变化过程详细写一下

[ sin(π/2 - x ) ] / [ 1 - cos(π/2 - x ) ]= [ sin2(π/4 - x/2 ) ] / [ 1 - cos2(π/4 - x/2 ) ] →2倍角公式→= [ 2sin(π/4 - x/2 )cos(π/4 - x/2 ) ] / { 1 - 1 + 2[ sin(π/4 - x/2 ) ]^2 }=[ 2sin(π/4 - x/2 )cos(π/4 - x/2 ) ] / {2[sin(π/4 - x/2 ) ]^2 }=[ cos(π/4 - x/2 ) ] / [sin(π/4 - x/2 ) ] 。

以上解答满意了么？

sin(½π-x)=sin[2(¼π-½x)]=2sin(¼π-½x)·cos(¼π-½x) 倍角公式1-cos(½π-x)=1-cos[2(¼π-½x)]=1-[1-2sin²(¼π-½x)]=2sin²(¼π-½x)倍角公式∴左边=2sin(¼π-½x)·cos(¼π-½x)/2sin²(¼π-½x) =cos(¼π-½x)/sin(¼π-½x) =cot(¼π-½x)

sin(π/2-x)/[1-cos(π/2-x)]=2sin(π/4-x/2)cos(π/4-x/2)/[2sin²(π/4-x/2)]=cos(π/4-x/2)/sin(π/4-x/2)