Question

Prove that: ((sinx^2)/(1+cotx))+((cosx^2)/(1+tanx))=1-cosxsinx Unsure of how to prove it?

Answer 1

`LHS=sin^2x/(1+cotx)+cos^2x/(1+tanx)`

`=sin^3x/(sinx(1+cotx))+cos^3x/(cosx(1+tanx))`

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

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

`=((sinx+cosx)(sin^2x-sinxcosx+cos^2x))/(cosx+sinx)`

`=sin^2x-sinxcosx+cos^2x`

`=1-sinxcosx=RHS`

Answer 2

Please refer to a Proof in the Explanation.

Explanation:

I hope, the Question is, to prove that,

`sin^2x/(1+cotx)+cos^2x/(1+tanx)=1-cosxsinx.`,#

We star with,

`"The L.H.S.="sin^2x/(1+cosx/sinx)+cos^2x/(1+sinx/cosx),`

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

`=sin^3x/(sinx+cosx)+cos^3x/(sinx+cosx),`

`=(sin^3x+cos^3x)/(sinx+cosx)={(sinx)^3+(cosx)^3}/(sinx+cosx).`

Recall that, `a^3+b^3=(a+b)(a^2-ab+b^2).` Therefore,

`"The L.H.S.="{cancel((sinx+cosx))(sin^2x-sinxcosx+cos^2x)}/cancel((sinx+cosx)),`

`=(sin^2x+cos^2x)-sinxcosx,`

`=1-sinxcosx,`

`"=The R.H.S."`

Hence, the Proof.

Enjoy Maths.!