Question

化简 tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2

Answer 1

用到两角和正切公式的变形

tan(M+N)=(tanM+tanN)/(1-tanMtanN)

∴ tanM+tanN=tan(M+N)*(1-tanMtanN)

∴ tanB/2tanA/2+tanB/2tanC/2+tanC/2tanA/2

= tan(B/2)[tan(A/2)+tan(C/2)]+tan(C/2)*tan(A/2)

=tan(B/2)*tan(A/2+C/2)*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)

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

=[sin(B/2)/cos(B/2)]*[sin(π/2-B/2)/cos(π/2-B/2)]*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)

=[sin(B/2)/cos(B/2)]*[cos(B/2)/sin(B/2)]*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)

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

=1-tan(C/2)tan(A/2)+tan(C/2)tan(A/2)

=1