Question

tanA(1+sec2A)=tan2A prove that:

Answer 1

LHS = tan A ( 1 + sec 2A )

.......= tan A ( 1 + 1/ cos 2A )

.......= tan A ( cos 2A + 1 ) / ( cos 2A )

.......= tan A ( 2 cos^2 A ) / ( cos 2A )

......= ( sin A / cos A ) ( 2 cos^2 A ) / ( cos 2A )

.......= ( 2 sin A cos A ) / cos 2A

.......= sin 2A / cos 2A

.......= tan 2A

........=RHS

Answer 2

TanA(1+Sec2A)

=TanA(1 + 1/Cos2A)

=TanA(1 + 1/Cos^2A-Sin^2A)

=TanA(Cos^2A-Sin^2A+Sin^2A+Cos^2A/Cos^2A-Sin^2A)

=SinA/Cos × (2Cos^2A/Cos^2A-Sin^2A)

=SinA × 2CosA/Cos2A

=2SinACosA/Cos2A

=Sin2A/Cos2A

=Tan2a

Step-by-step explanation:In the second line.. We know Sec2A=1/Cos2A(As sec is opposite to Cos). In the third line due to a rule Cos2A=Cos^2A-Sin^2A.Then by doing L. C. M. In the next line we divide TanA to SinA/CosA. Previous than the final line by dividing 2SinACosA and Cos2A.....We get Sin2A/Cos2A. And it results Tan2A........ Hope you all viewers could understood it.