# 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 =...

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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

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