Prove that: ^2A+ ^2B-2 (A-B)= ^2(A+B)

Question

Prove that:$\cos^2\text{A}+\cos^2\text{B}-2\cos\text{A}\cos\text{B}\cos\text{(A}-\text{B)}=\sin^2\text{(A}+\text{B)} $

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Answer

$\text{R.H.S}=\cos^2\text{A}+\cos^2\text{B}-2\cos\text{A}\cos\text{B}\cos\text{(A}+\text{B)}$
$=\cos^2\text{A}+(\text{1}-\sin^2\text{B})-2\cos\text{A}\cos\text{B}\cos(\text{A}+\text{B})$
$=\Big[\cos^2\text{A}-\sin^2\text{B}\Big]-2\cos\text{A}\cos\text{B}\cos\text{(A}+\text{B)}+1$
$=\Big[\cos\text{(A}+\text{B)}\cos\text{(A}-\text{B)}\Big]-2\cos\text{A}\cos\text{B}+\cos\text{(A}+\text{B}) +1$
$=\cos\text{(A}+\text{B)}\Big[\cos(\text{A}-\text{B)}-2\cos\text{A}\cos\text{B}\Big] +1$
$=\cos\text{(A}+\text{B)}\Big[\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}-2\cos\text{A}\cos\text{B}\Big]+1$
$=\cos\text{(A}+\text{B)}\Big[-\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}\Big]+1$
$=\cos\text{(A}+\text{B)}\Big[\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}\Big]+1$
$ =\cos^2\text{(A}+\text{B)}+1$
$=1-\cos^2\text{(A}+\text{B)}$ $\Big[\sin^2\theta=1-\cos^2\theta\Big]$
$=\sin^2\text{(A}+\text{B)}$
$=\text{R.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
Hence proved.

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