show that sin2A÷1+cos2A=tan A
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: sin2A=2sinAcosA ; cos2A=2cos²A –1
LHS;
=2sinAcosA÷1+2cos²–1
=sinAcosA÷cosA×cosA
=sinA÷cosA { sinA÷cosA = tanA}
=tanA
Hence, LHS=RHS
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