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mark me as brainliest. hope this helps
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to prove sinA-2sin^3A÷2cos^3A-cosA=tanA
now
sinA (1-2sin^2A)÷cosA (2cos^2A-1)
we can write
sinA÷cosA (1-2sin^2A÷2cos^2 A-1)
as we know
sinA÷csA=tanA
therefore
tanA (1-2sin^2 A ÷2cosA^2A-1)
tanA (1-sin^2A-sin^2A÷cos^2A+cos^2-1)
as we know that 1-sin^2A=cos^2A
and cos^2A-1=-sin^2A
therefore
tanA (cos^2-sin^2A÷cos^2A-sin^2A)
tanA (proove )
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