Prove that: ^2(n+1)A- ^2nA= (2n+1)A - Vidyadip

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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Answer

We have,
$\text{L.H.S} =\sin^2\text{(n+1)}\text{A}-\sin^2\text{nA}$
$=\sin[(\text{n+1)}\text{A}+\text{nA}]\sin[(\text{n+1)}\text{A}-\text{nA}]$
$\Big[\because\sin^2\text{A}-\sin^2\text{B}=\sin\text{(A+ B)}\sin\text{(A-B)}\Big]$
$=\sin[\text{nA}+\text{A}+\text{nA}]\sin[\text{nA}+\text{A}-\text{nA}]$
$=\sin\text{(2nA+A)}\sin\text{(A)}$
$=\sin\text{(2n+1)}\text{A}\sin\text{A}$
$=\text{R.H.S}$
$$$\therefore\text{L.H.S}=\text{R.H.S}$
Hence proved.

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