Prove that (1+tan²A) cosA sinA =tanA
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We know , 1+tan²A = sec²A = 1/cos²A
(1/cos²A) × sinAcosA = sinA/cosA = tanA
(1/cos²A) × sinAcosA = sinA/cosA = tanA
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LHS=(1+tan^2A)cotAsinA
=sec^2A×cotAsinA
=1÷cos^2A×cosAsinA
=sinA÷cosA
=tanA=RHS
=sec^2A×cotAsinA
=1÷cos^2A×cosAsinA
=sinA÷cosA
=tanA=RHS
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