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If sinA=sinBa n dcosA=cosB , then prove ...

If `sinA=sinBa n dcosA=cosB ,` then prove that `sin(A-B)/2=0`

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`sinA = sin^2B ->(1)`
`2cos^2A = 3cos^2B`
`=>2cos^2A = 3(1-sin^2B)`
From (1),
`=>2(1-sin^2A) = 3(1-sinA)`
`=>2sin^2A - 3sinA+1 = 0`
`=>2sin^2A - 2sinA-sinA+1 = 0`
`=>2sinA(sinA-1)-1(sinA-1) = 0`
...
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