已知△ABC中,角A,B,C所对的边分别是a,b,c,且a+c=√2b

1)正弦定理 a/sinA=b/sinB=c/sinC ∴sinA+sinC=√2sinB=√2sin(180°-A-C)=√2sin(A+C) ∴2sin(A+C)/2cos(A-C)/2=√2×2sin(A+C)/2cos(A+C)/2 ∴cos(A-C)/2=√2cos(A+C)/2 ∴cosA/2cosC/2+sinA/2sinC/2=√2cosA/2cosC/2-√2sinA/2sinC/2 ∴(√2-1)cosA/2cosC/2=(√2+1)sinA/2sinC/2 ∴tanA/2tanC/2=sinA/2sinB/2 / cosA/2cosC/2=(√2-1)/(√2+1)=(√2-1)=3-2√2 2)1/tanA+1/tanC=(1-tanA/2)/2tanA/2+(1-tanC/2)/2tanC/2 =(tanA/2+tanC/2-tanA/2tanC/2-tanC/2tanA/2)/2tanA/2tanC/2 =(tanA/2+tanC/2)(1-tanA/2tanC/2)/2tanA/2tanC/2 =(tanA/2+tanC/2)(1-3+2√2)/2(3-2√2) =(tanA/2+tanC/2)/(3-2√2)(√2+1) =(tanA/2+tanC/2)/ (√2-1) =2(tanA/2+tanC/2)/ [1-(3-2√2)] =2(tanA/2+tanC/2)/(1-tanA/2tanC/2) =2tan(A+C)/2=2cotB/2=2/tanB/2