How can I prove that " 1 - (sin^2x / 1 + cotx) - (cos^2x / 1 + tanx) = sinxcosx " using trig identities?

We need to prove that:

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

We will start from
the left side and prove that it equals the right side.

Let us
preview some trigonometric properties:

We know that
:

tanx = sinx/cosx

cotx = cosx/
sinx

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

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

==> ( 1- ( sin^3 x)/ (sinx+ cosx) - (
cos^3 x/ (cosx+ sinx)

==> (sinx+cosx - sin^3 x - cos^3 x) /
(cosx + sinx)

We will combine similar terms
together.

==> (sinx-sinx^3 x + cosx - cos^3 x)/(cosx +
sinx)

Now we will factor sinx from the first 2 terms and cosx from
the last two terms.

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

Now we know that: sin^2 x + cos^2 x =
1

==> 1-sin^2 x = cos^2
x

==> 1- cos^2 x = sin^2
x

==> (sinx*cos^2x + cosx*sin^2x)/
(cosx+sinx)

We will factor
cosx*sinx

==> cosx*sinx( cosx + sinx)/(cosx +
sinx)

==> cosx*
sinx................q.e.d