Question

2sinA.cosA=2tanA/1+tan squareA

Answer 1

l.h.s= sin2A = sin(A+A)

= sinA.cosA+cosA.sinA

=2SinA.cosA----------------we call 1st eqn

[...1=cos2A+sin2A is a formula]

We assume that 2sinA.cosA/1

=2sinA.cosA/sin2A+cos2A

Let take'cos2A' Dividing on both numinator and denominator by 'cos2A'

=2sinA.CosA/cos2A/cos2A+sin2A/Cos2A

=2sinA/cosA/cos2A/cos2A + Sin2A/cos2A [sinA/cosA=tanA] [sin2A/cos2A=tan2A]

=2tanA/1+tan2A

L.H.S=R.H.S

Answer 2

Take RHS

2TanA/(1+Tan^2A)

2SinA/CosA /(1+Sin^2A/Cos^2A)

2SinA/CosA /(Cos^2A+Sin^2A/Cos^2A )

2SinA/CosA/(1/Cos^2A)

2SinACos^2A/CosA

2SinACosA = LHS

Hence proved