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

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To  prove 1 - (sin^2x / 1 + cotx) - (cos^2x / 1 + tanx) = sinxcosx .

We use cot x= cosx/sinx and tanx = sinx/cosx in LHS. as the right side as terms sinx /cosx.

LHS:

1 - (sin^2x / 1 + cotx) - (cos^2x / 1 + tanx).

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

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

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

= 1-(sin^2x-sixcosx+cos^2x) , as (a^3+b^3)/(a+b) = (a^2-ab+b^2).

= 1 - (sin^2+cos^2x-sinxcosx).

1-(1-sinxcosx), as sin^2+cos^2x = 1 is a trigonometric identity.

1-1+sinxcosx.

= sinxcosx = RHS.

Therefore ,

1 - (sin^2x / 1 + cotx) - (cos^2x / 1 + tanx) = sinxcosx.

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

L:H:S ≡ 1- sin²x/( 1+ cotx) - cos²x/ (1+ tanx)

= 1 - sin²x÷[(sinx+cosx)/sinx] - cos²x÷[(cosx+sinx)/cosx]

= 1 - sin³x/(sinx+cosx) - cos³x/(cosx+sinx)

= 1 - [1/(sinx+cosx)]*[sin³x+cos³x]

⇒ x³ + y³ = (x + y)(x² + y² - xy)

= 1 - [1/(sinx+cosx)]*[(sinx+cosx)(sin²x+cos²x-sinx.cosx)]

= 1- (sin²x+cos²x-sinx.cosx)

we know that, sin²x+cos²x=1

= 1 - (1-sinx.cosx)

= sinx.cosx

=R:H:S