`sin(2n+1)A.sinA = sin^2(n+1)A-sin^2(nA)`

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Prove that sin^2(n+1)A-sin^2nA=sin(2n+1)AsinA

Prove that: sin^(2)(n+1)A-sin^(2)nA=sin(2n+1)As in A

Prove that: sin^2(n+1)A-sin^2n A="sin"(2n+1)Asin A

sin A+sin3A+...+sin(2n-1)A=(sin^(2)nA)/(sin A)

(sin (n +1) A +2 sin n A + sin ( n -1) A )/( cos (n +1) A - cos ( n -1) A ) =

(sin (n +1) A + 2 sin n A + sin (n-1) A)/( cos ( n -1) A - cos (n +1) A)=

(sin (n + 1) A + 2sin n + sin (n-1) A) / (cos (n-1) A-cos (n + 1) A) = (cot A) / (2)

sin^(2)n theta-sin^(2)(n-1)theta=sin^(2)theta where n is constant and n!=0,1

sin (n + 1) A * sin (n-1) A + cos (n + 1) A * cos (n-1) A = cos2A

(Show sin (n + 1) A-sin (n-1) A) / (cos (n + 1) A + 2cos nA + cos (n-1) A) = tan ((A) / (2))