Question
Prove that: $\sin^2\text{(n+1)}\text{A}-\sin^2\text{nA}=\sin\text{(2n+1)}\text{A}\sin\text{A}$
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units.
$\frac{n^2-9}{(n+3) !}+\frac{6}{(n+2) !}-\frac{1}{(n+1) !}$