# Proving trigonometric identities Prove: sin2A = 2tanA/1 + tan^2A You want  the following to be proved: sin 2A = 2tan A/ (1 + (tan A)^2).

We know that tan A  = sin A/ cos A and (sin A)^2 + (cos A)^2 = 1.

2tan A/ (1 + (tan A)^2)

=> [2* sin A / cos A]/ [ 1 + (sin A)^2 / (cos A)^2]

=> [2* sin A / cos A]/ [ ((cos A)^2 + (sin A)^2) / (cos A)^2]

=> [2 sin A* (cos A)^2 / cos A]/ [ (cos A)^2 + (sin A)^2]

=> [2 sin A * cos A] / 1

=> sin 2A

Therefore sin 2A = 2tan A/ (1 + (tan A)^2).

Approved by eNotes Editorial Team We need to prove that:

sin2A = 2tanA/ (1+tan^2 A)

We will start from the right side and prove the left side.

We know that tanA = sinA/cosA

==> 2tanA / (1+tan^2 A) = 2(sinA/cosA) / [1+ (sinA/cosA)^2]

= 2sinA/ cosA[ 1+ (sin^2A/cos^2 A)]

= 2sinA/ cosA*[(cos^2 A + sin^2 A)/cos^2A]

But we know that sin^2 A + cos^2 A = 1

==> 2tanA/ (1+tan^2 A) = 2sinA/  (1/cosA)

= 2sinA*cosA

But we know that sin2A = 2sinA*cosA

==> 2tanA / (1+tan^2 A) = sin2A

Approved by eNotes Editorial Team