Prove that Sin theta - cos theta whole square = 1-sin square theta
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Answered by
0
Answer:
I THINK YOU HAVE WRITTEN WRONG QUESTION
Step-by-step explanation:
(SINA-COSA)^2=1-SIN2A
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Answered by
1
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.
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