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Miscellaneous Exercise 3 (I) · Q86

Q.Select the correct option: Let 0<A,B<π20<A,B<\dfrac{\pi}{2} satisfying the equations 3sin⁡2A+2sin⁡2B=13\sin^2A+2\sin^2B=1 and 3sin⁡2A−2sin⁡2B=03\sin2A-2\sin2B=0; then A+2BA+2B is equal to (A) π\pi (B) π2\dfrac{\pi}{2} (C) π4\dfrac{\pi}{4} (D) 2π2\pi

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Step 1: From 3sin⁡2A+2sin⁡2B=13\sin^2A+2\sin^2B=1: 3sin⁡2A=1−2sin⁡2B=cos⁡2B3\sin^2A=1-2\sin^2B=\cos2B (using cos⁡2B=1−2sin⁡2B\cos2B=1-2\sin^2B). Call this (I).

Step 2: From 3sin⁡2A=2sin⁡2B3\sin2A=2\sin2B: sin⁡2B=32sin⁡2A=3sin⁡Acos⁡A\sin2B=\dfrac32\sin2A=3\sin A\cos A. Call this (II).

Step 3: Compute cos⁡(A+2B)=cos⁡Acos⁡2B−sin⁡Asin⁡2B\cos(A+2B)=\cos A\cos2B-\sin A\sin2B; substitute (I) and (II): =cos⁡A(3sin⁡2A)−sin⁡A(3sin⁡Acos⁡A)=3sin⁡2Acos⁡A−3sin⁡2Acos⁡A=0=\cos A(3\sin^2A)-\sin A(3\sin A\cos A)=3\sin^2A\cos A-3\sin^2A\cos A=0. …

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