Prove that (1-cos𝛉) (1+cos𝛉) =sin2(square)𝛉

 LHS=(1-cos𝛉) (1+cos𝛉) 

           1(1+cos𝛉) -cos𝛉(1+cos𝛉) 

            1+cos𝛉-cos𝛉-cos2(square)𝛉

              1-cos2(square)

               =sin2(square)𝛉

2nd Method

(1-cos𝛉) (1+cos𝛉) 

let a=1 and b=cos𝛉

know (a-b) (a+b)

=a2(square)-b2(square)

=(1)(square)-cos2(square)𝛉

=sin2(square)𝛉