CẦN:• Ta có
÷a,b,c⇔a+c=2b
⇔2RsinA+2RsinC=2.2RsinB
⇔sinA+sinC=2sinB
⇔2sinA+C2cosA−C2=4sinB2cosB2=4sinA+C2cosA+C2
⇔cosA−C2=2cosA+C2(A+C2≠0,A+C2≠π)
⇔cosA2cosC2+sinA2sinC2=2cosA2cosC2−2sinA2sinC2
⇔3sinA2sinC2=cosA2cosC2
⇔3tanA2tanC2=1
⇔tanA2tanC2=13.
ĐỦ:• Ta có
tanA2tanC2=13
⇔3tanA2tanC2=1
⇔3sinA2sinC2=cosA2cosC2
⇔cosA2cosC2+sinA2sinC2=2cosA2cosC2−2sinA2sinC2
⇔cosA−C2=2cosA+C2(A+C2≠0,A+C2≠π)
⇔2sinA+C2cosA−C2=4sinB2cosB2=4sinA+C2cosA+C2
⇔sinA+sinC=2sinB
⇔2RsinA+2RsinC=2.2RsinB
⇔a+c=2b.
Vậy ta có đpcm.