prove that sin square A+sin square (A-B) - 2sinAcosBsin(A-B)=sin square B
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the problem is solved......
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tep-by-step explanation:
Consider the provided information.
\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B
Consider the LHS.
\sin^2A\cos^2B-\cos^2A\sin^2B
\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B (∴\cos^2x=1-\sin^2x)
\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B
\sin^2A-\sin^2B
Hence, proved.
Step-by-step explanation:
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