You can use the substitution rules based on the Eq.8.126 from Table Of Integrals, Series And Products by Gradshteyn and Ryzhik. I retype them below in original notations $$K\!\left(\frac{2\sqrt{k}}{1+k}\right)=(1+k)K(k),$$ and $$E\!\left(\frac{2\sqrt{k}}{1+k}\right)=\frac1{1+k}\left(2E(k)-(1-k^2)K(k)\right).$$ One should remember that they are valid for $0\le k\le 1$ and one should beware of slightly different MA conventions. They are called quadratic arithmetic-geometric mean (AGM) transformations for the complete elliptic integrals.

rules={EllipticK[(4 λ_)/(1+λ_)^2]-> (1+λ)EllipticK[λ^2], 
EllipticE[(4 λ_)/(1+λ_)^2]-> 1/(1+λ) (2EllipticE[λ^2]-(1-λ^2)EllipticK[λ^2])};

Now with

f1[a_] = (2/π) Integrate[Cos[t]/(1 - 2 a Cos[t] + a^2)^(3/2), {t, 0, π}, 
Assumptions -> {0 < a < 1}];
f2[a_] = (4/(π a (1 - a^2)^2)) ((1 + a^2) EllipticE[a^2] -
        (1 - a^2)EllipticK[a^2]);

Then

Simplify[(f1[a] /. rules) - f2[a]]

returns

(* 0 *)

You can use the substitution rules based on the Eq.8.126 from Table Of Integrals, Series And Products by Gradshteyn and Ryzhik. They are called quadratic arithmetic-geometric mean (AGM) transformations for the complete elliptic integrals. I retype them below in original notations $$K\!\left(\frac{2\sqrt{k}}{1+k}\right)=(1+k)K(k),$$ and $$E\!\left(\frac{2\sqrt{k}}{1+k}\right)=\frac1{1+k}\left(2E(k)-(1-k^2)K(k)\right).$$ One should remember that they are valid for $0\le k\le 1$ and one should beware of slightly different MA conventions.

rules={EllipticK[(4 λ_)/(1+λ_)^2]-> (1+λ)EllipticK[λ^2], 
EllipticE[(4 λ_)/(1+λ_)^2]-> 1/(1+λ) (2EllipticE[λ^2]-(1-λ^2)EllipticK[λ^2])};

Now with

f1[a_] = (2/π) Integrate[Cos[t]/(1 - 2 a Cos[t] + a^2)^(3/2), {t, 0, π}, 
Assumptions -> {0 < a < 1}];
f2[a_] = (4/(π a (1 - a^2)^2)) ((1 + a^2) EllipticE[a^2] -
        (1 - a^2)EllipticK[a^2]);

Then

Simplify[(f1[a] /. rules) - f2[a]]

returns

(* 0 *)

You can use the substitution rules based on the Eq.8.126 from Table Of Integrals, Series And Products by Gradshteyn and Ryzhik. I retype them below in original notations $$K\!\left(\frac{2\sqrt{k}}{1+k}\right)=(1+k)K(k),$$ and $$E\!\left(\frac{2\sqrt{k}}{1+k}\right)=\frac1{1+k}\left(2E(k)-(1-k^2)K(k)\right).$$ One should remember that they are valid for $0\le k\le 1$ and one should beware of slightly different MA conventions. They are called quadratic arithmetic-geometric mean (AGM) transformations for the complete elliptic integrals.

rules={EllipticK[(4 λ_)/(1+λ_)^2]-> (1+λ)EllipticK[λ^2], EllipticE[(4 λ_)/(1+λ_)^2]-> 1/(1+λ) (2EllipticE[λ^2]-(1-λ^2)EllipticK[λ^2])};
f1[a_] = (2/π) Integrate[Cos[t]/(1 - 2 a Cos[t] + a^2)^(3/2), {t, 0, π}, Assumptions -> {0 < a < 1}];
f2[a_] = (4/(π a (1 - a^2)^2)) ((1 + a^2) EllipticE[a^2] - (1 - a^2)EllipticK[a^2]);
Simplify[(f1[a] /. rules) - f2[a]]
(*0*)

You can use the substitution rules based on the Eq.8.126 from Table Of Integrals, Series And Products by Gradshteyn and Ryzhik. I retype them below in original notations $$K\!\left(\frac{2\sqrt{k}}{1+k}\right)=(1+k)K(k),$$ and $$E\!\left(\frac{2\sqrt{k}}{1+k}\right)=\frac1{1+k}\left(2E(k)-(1-k^2)K(k)\right).$$ One should remember that they are valid for $0\le k\le 1$ and one should beware of slightly different MA conventions. They are called quadratic arithmetic-geometric mean (AGM) transformations for the complete elliptic integrals.

rules={EllipticK[(4 λ_)/(1+λ_)^2]-> (1+λ)EllipticK[λ^2],  
EllipticE[(4 λ_)/(1+λ_)^2]-> 1/(1+λ) (2EllipticE[λ^2]-(1-λ^2)EllipticK[λ^2])};

Now with

f1[a_] = (2/π) Integrate[Cos[t]/(1 - 2 a Cos[t] + a^2)^(3/2), {t, 0, π},  
Assumptions -> {0 < a < 1}];
f2[a_] = (4/(π a (1 - a^2)^2)) ((1 + a^2) EllipticE[a^2] -
        (1 - a^2)EllipticK[a^2]);

Then

Simplify[(f1[a] /. rules) - f2[a]]

returns

(* 0 *)