Question

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

Answer 1

`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`