Question
Prove the following trigonometric identities.
$\sin^2\text{A}\cos^2\text{B}-\cos^2\text{A}\sin^2\text{B}=\sin^2\text{A}-\sin^2\text{B}$
$\sin^2\text{A}\cos^2\text{B}-\cos^2\text{A}\sin^2\text{B}=\sin^2\text{A}-\sin^2\text{B}$
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Answer
$\text{L.H.S}=\sin^2\text{A}\cos^2\text{B}-\cos^2\text{A}\sin^2\text{B}$
$=\sin^2\text{A}(1-\sin^2\text{B})-(1-\sin^2\text{A})(\sin^2\text{B})$
$(\because \cos^2\text{A}=1-\sin^2\text{A})$
$=\sin^2\text{A}-\sin^2\text{A}\sin^2\text{B}-\sin^2\text{B}+\sin^2\text{A}\sin^2\text{B}$
$=\sin^2\text{A}-\sin^2\text{B}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$=\sin^2\text{A}(1-\sin^2\text{B})-(1-\sin^2\text{A})(\sin^2\text{B})$
$(\because \cos^2\text{A}=1-\sin^2\text{A})$
$=\sin^2\text{A}-\sin^2\text{A}\sin^2\text{B}-\sin^2\text{B}+\sin^2\text{A}\sin^2\text{B}$
$=\sin^2\text{A}-\sin^2\text{B}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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