Prove the following identities: ( - ) ( + ) + ( - ) ( + ) =0

Question

Prove the following identities:
$\frac{(\sin\text{A}-\sin\text{B})}{(\cos\text{A}+\cos\text{B})}+\frac{(\cos\text{A}-\cos\text{B})}{(\sin\text{A}+\sin\text{B})}=0$

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Answer

$\text{LHS}=\frac{(\sin\text{A}-\sin\text{B})}{(\cos\text{A}+\cos\text{B})}+\frac{(\cos\text{A}-\cos\text{B})}{(\sin\text{A}+\sin\text{B})}$
$=\frac{(\sin\text{A}+\sin\text{B})(\sin\text{A}-\sin\text{B})+(\cos\text{A}+\cos\text{B})(\cos\text{A}-\cos\text{B})}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{\sin^2\text{A}-\sin^2\text{B}+\cos^2\text{A}-\cos^2\text{B}}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{\big(\sin^2\text{A}+\cos^2\text{A}\big)-\big(\sin^2\text{A}+\cos^2\text{B}\big)}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{1-1}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=0$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$

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