We will start from the right side to the left.

We will use trigonometric identities to prove the identity.

==> 2sinxcosx/(1+ cos^2 x- sin62 x) = tanx.

We know that sin^2 x + cos^2 x = 1

==> 2sinxcosx/ (sin^2 x+cos^2 + cos^2 x-sin62 x)

==> 2sinxcosx / 2cos^2 x

Now we will reduce similar terms.

==> sinx/ cosx = tanx.

**Then, we have proved that tanx = 2sinx*cosx/ (1+cos^2 x- sin^2x).**

We have to prove that tanx = 2sinx*cosx/(1+cos^2x-sin^2x)

Let's start from the right.

2sinx*cosx/(1+cos^2x-sin^2x)

=> 2*sin x * cos x /( 1 + (cos x)^2 - (sin x)^2)

=> 2*sin x * cos x /( 1 + (cos x)^2 - 1 + cos x)^2)

=> 2*sin x * cos x / 2*(cos x)^2

=>sin x / cos x

=> tan x

which is the left hand side.

**This proves that : tanx = 2sinx*cosx/(1+cos^2x-sin^2x)**

We'll replace 1 from denominator of the fraction by Pythagorean identity:

(sin x)^2 + (cos x)^2 = 1

The denominator will become:

(sin x)^2 + (cos x)^2 + (cos x)^2 - (sin x)^2 = 2 (cos x)^2

We'll re-write the fraction:

2sin x*cos x/2 (cos x)^2 = sin x/cos x

But the fraction sin x/cos x represents the tangent function.

**Since the LHS = RHS, then the given identity tan x = 2 sinx*cos x/[1 + (cos x)^2 - (sin x)^2] is true.**