Question

prove that
1+tan2A.tanA=sec2A

Answer 1

We know that Tan2A = Sin2A/Cos2A and TanA = SinA/CosA

substituting we get

$\frac{Cos2A CosA + Sin2A SinA}{Cos2A CosA}$ = 1 + Tan2ATanA

but Cos(A - B) = CosACosB +SinASinB

⇒Cos2A CosA + Sin2A SinA = Cos(2A - A) = CosA

⇒ $\frac{CosA}{Cos2A CosA}$ = 1 + Tan2ATanA

⇒ 1/Cos2A = 1 + Tan2ATanA

⇒ Sec2A = 1 + Tan2ATanA