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Miscellaneous Exercise 2 · Q78

Q.Prove that sin⁡2Acos⁡2B+cos⁡2Asin⁡2B+cos⁡2Acos⁡2B+sin⁡2Asin⁡2B=1\sin^2 A\cos^2 B + \cos^2 A\sin^2 B + \cos^2 A\cos^2 B + \sin^2 A\sin^2 B = 1.

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Step 1. Group: (sin⁡2Acos⁡2B+cos⁡2Acos⁡2B)+(cos⁡2Asin⁡2B+sin⁡2Asin⁡2B)(\sin^2A\cos^2B+\cos^2A\cos^2B) + (\cos^2A\sin^2B+\sin^2A\sin^2B).

Step 2. Factor: cos⁡2B(sin⁡2A+cos⁡2A)+sin⁡2B(cos⁡2A+sin⁡2A)=cos⁡2B(1)+sin⁡2B(1)\cos^2B(\sin^2A+\cos^2A) + \sin^2B(\cos^2A+\sin^2A) = \cos^2B(1)+\sin^2B(1). …

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