求助高中数学一个三角函数问题

证明：
由于：A属于(0,π/2)
则：sinA>0,cosA>0
则：
tanA+√[(tanA)^2+1]

=tanA+√[(sinA)^2/(cosA)^2 +1]

=tanA+√{[(sinA)^2+(cosA)^2]/(cosA)^2}

=tanA+√[1/(cosA)^2]

=tanA+(1/cosA)

=sinA/cosA+(1/cosA)

=(sinA+1)/cosA

=[2sin(A/2)cos(A/2)+1]/[cos^2(A/2)-sin^2(A/2)](二倍角公式)

=[2sin(A/2)cos(A/2)+sin^2(A/2)+cos^2(A/2)]
/[cos^2(A/2)-sin^2(A/2)]
(将1化为正余弦平方和的形式)

=[sin(A/2)+cos(A/2)]^2/{[cos(A/2)-sin(A/2)][cos(A/2)+sin(A/2)]}
(因式分解)

=[sin(A/2)+cos(A/2)]/[cos(A/2)-sin(A/2)]
上下同时除以cos(A/2)得：
=[tan(A/2)+1]/[1-tan(A/2)]

=[tan(A/2)+tan(π/4)]/[1-tan(π/4)*tan(A/2)]

=tan(π/4+A/2)

tan A+(tan^2 A+1)^0.5=tan A+secA
=(sinA+1)/cosA
=[cos(A/2)+sin(A/2)]^2/[cos^2(A/2)-sin^2(A/2)
=[cos(A/2)+sin(A/2)]/[cos(A/2)-sin(A/2)]
=[1+tan(A/2)]/[1-tan(A/2)]
=tan(π/4+A/2)

tanπ/4=1