prove that if in triangle ABC tan(A/2)=5/6 tan(B/2)=2/5 then a,b,c will be in AP
$\displaystyle \displaystyle\ tan\left[\frac{A}{2}\right]=\frac{5}{6}\Rightarrow\ sinA=\frac{2tan\left[\frac{A}{2}\right]}{1+tan^2\left[\frac{A}{2}\right]}=\frac{\left[\frac{5}{3}\right]}{\left[\frac{61}{36}\right]}=\frac{60}{61}$
$\displaystyle \displaystyle\ tan\left[\frac{B}{2}\right]=\frac{2}{5}\Rightarrow\ sinB=\frac{\left[\frac{4}{5}\right]}{\left[\frac{29}{25}\right]}=\frac{20}{29}$
$\displaystyle sinC=sin[180^o-(A+B)]=sin(A+B)=sinAcosB+cosAsinB$
We obtain the cosines from Pythagoras' Theorem.
$\displaystyle x^2+60^2=61^2\Rightarrow\ x^2=3721-3600=121\Rightarrow\ x=11$
$\displaystyle \displaystyle\Rightarrow\ cosA=\frac{11}{61}$
$\displaystyle y^2+20^2=29^2\Rightarrow\ y^2=29^2-20^2=441\Rightarrow\ y=21$
$\displaystyle \displaystyle\Rightarrow\ cosB=\frac{21}{29}$
$\displaystyle \displaystyle\ sinC=\frac{60}{61}\;\frac{21}{29}+\frac{11}{61}\;\ frac{20}{29}=\frac{1260+220}{1769}=\frac{1480}{176 9}$
From the Law of Sines
$\displaystyle \displaystyle\frac{a}{sinA}=\frac{b}{sinB}=\frac{c }{sinC}$
$\displaystyle \Rightarrow\ a:b:c=sinA:sinB:sinC$
If a, b, and c are in Arithmetic progression,
then the sum of the 3 sides is 3 times the "middle side".
$\displaystyle p+(p+d)+(p+2d)=3p+3d=3(p+d)$
Hence, the sum of the 3 sines must be 3 times the sum of the "middle sine".
$\displaystyle sinA+sinB+sinC=2.50989259469$
$\displaystyle 3sinC=2.50989259469$
Therefore
a, c, b are in Arithmetic Progression.
b, c, a are in Arithmetic Progression.