prove that:sin2(n + 1)A - sin2nA = sin(2n + 1)A.sinA

sin²(n+1)A - sin²nA = sin(2n+1)A.sinA

we have formula sin²a - sin²b =sin(a-b)sin(a+b)

therefore

                 sin²(n+1)A - sin²nA=sin[(n+1)A+nA].sin[(n+1)A-nA]

                                            =sin(2n+1)A.sinA =RHS

hence proved

Approved

sin^2(n + 1)A - sin^2nA = sin(2n + 1)A.sinA
Solution:-
    L.H.S:-sin^2(n + 1)A - sin^2nA =
        Sin[(n+1)A+nA].Sin[(n+1)A-nA]        ["a^2-b^2=(a+b)(a-b)"]
        =Sin[nA+A+nA].Sin[nA+A-nA]
        =Sin[A(n+1+n)].SinA
        =Sin(2n+1)A.SinA=R.H.S

Thank You!!!!!!!!!!!!!!!!

Wish You Happy learning of trignometry..............

Approved