# Proving trigonometric identities Prove: sin2A = 2tanA/1 + tan^2A

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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**

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).**

L:H:S ≡ sin2A

**⇒ use sin 2A = 2sinA.cosA**

= 2sinA.cosA/1

**⇒ Divide top and bottom by cos²A**

= (2sinA.cosA/cos²A)/ (1/cos²A)

= (2sinA/cosA) / sec²A

**⇒ use sec²A = 1 + tan²A **

= 2tanA / 1+ tan²A

L:H:S ≡ R:H:S

in2A = 2tanA/(1+tan^2A).

We know that sin(A+B) = sinAcosB+cosAsinB is an identity. If we put B = A in this, we get:

sin(A+A) = sinAcosA+cosAsinA .

sin2A = 2sinAcosA.

We divide both sides by 1:

(sin2A)/1 = 2sinAcosA/(cos^2A+sin^2A) , as cos^2A+sin^2A) = 1.

sin2A = 2(sinAcosA/cos^2A)/{(sin^2A/cos^A +cos^2/cos^2A)}

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