Prove that Sin A -2sin cube A ÷2 cos cube A -cosA =tan A
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sin A -2sin3A /2cos3A-cosA
= sinA (1-2sin2A )/cosA ( 2cos2A - 1)
= sinA (1-2sin2A)/ cosA (2 (1-sin2A) -1)
=sinA (1-2 sin 2A)/cosA (1-2sin2A)
=sinA/cosA
=tanA
Hence proved
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= sinA (1-2sin2A )/cosA ( 2cos2A - 1)
= sinA (1-2sin2A)/ cosA (2 (1-sin2A) -1)
=sinA (1-2 sin 2A)/cosA (1-2sin2A)
=sinA/cosA
=tanA
Hence proved
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