求证tanx+1/tan[(π/4)+X/2]=1/COSX

tanx+1/tan[(π/4)+x/2]\
= tanx +1/[(tanπ/4+tanx/2)/(1-tanπ/4tanx/2)]
= tanx + 1/[(1+tanx/2)/(1-tanx/2)]
=tanx + (1-tanx/2)/(1+tanx/2)
= 2tanx/2/(1-tanx/2) + (1-tanx/2)/(1+tanx/2)
= [2tanx/2(1+tanx/2) + (1-tanx/2)^2]/ ( 1-(tanx/2)^2)
=[(tanx/2)^2+1] /( 1-(tanx/2)^2)
= (secx/2)^2 (cosx/2)^2/[(cosx/2)^2 - (sinx/2)^2]
= 1/[(cosx/2)^2 - (sinx/2)^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.