How To Prove ? TanA(1+Sec2A)=Tan2A

1 Answer
May 21, 2018

LHS=tanA(1+sec2A)

=tanA(1+1/(cos2A))=tanA((1+cos2A)/(cos2A))

=1/(cos2A)[tanA*2cos^2A]

=1/(cos2A)[sinA/cosA*2cos^2A]

=(sin2A)/(cos2A)=RHS