求证tanx+1/tan[(π/4)+X/2]=1/COSX

tan[(π/4)+X/2]= (tanπ/4+tan X/2)/(1- tanπ/4*tan X/2)
=(1+ tan X/2)/(1- tan X/2)
分子分母同乘以cosx/2可得
=(cosx/2+sinx/2)/( cosx/2-sinx/2)
=[(cosx/2+sinx/2) (cosx/2-sinx/2)]/( cosx/2-sinx/2) ²
=(cos²x/2-sin²x/2) /( cosx/2-sinx/2) ²
=cosx/(1-sinx),
所以1/tan[(π/4)+X/2]= (1-sinx)/cosx,
tanx+1/tan[(π/4)+X/2]= tanx+(1-sinx)/cosx
=sinx/cosx+(1-sinx)/cosx=1/cosx,
∴等式成立.