Question

2 tan A/1+tan^2A=sin2A​

Answer 1

Answer:

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 $tanA = sinA/cosA$

OR..........

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

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

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

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