Another Trigonometry identity question

Show that;
sin^3x+cos^3x/sinx+cosx = 1-sinxcosx

How do you do this?

Choosing the left side (Note the sum of two cubes in the numerator):

1. Factoring out (sinx + cosx) from the numerator gives: (sinx+cosx)(sin^2x-sinxcosx+cos^2x) / denom.

2. (sinx+cosx) and the denom cancel out, leaving: sin^2x-sinxcosx+cos^2x.

3. Being that sin^2x + cos^2x =1, you are left with 1-sinxcosx. QED

The given expression is ;

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

On re-arranging we have ;

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

Now on operating on the L.H.S we have ;

( sin^3x+cos^3x + sin^2x.cos x + cos^2x.sin x ) / ( sin x+cos x ) ;

= [ sin^2x.( sin x + cos x) + cos^2x.( sin x + cos x ) ] / ( sin x+cos x ) ..[taking sin^2x and
and cos^2x common ]

= [ ( sin^2x + cos^2x).(sin x + cos x ) ] / ( sin x+cos x )

= [ 1 . ( sin x+cos x ) ] / ( sin x+cos x ) ...[Since sin^2x + cos^2x = 1 ]

= ( sin x+cos x ) / ( sin x+cos x ) = 1 = R.H.S ...[Hence proved]

(a^3 + b^3) = (a + b) * (a^2 - ab + b^2)

(sin(x)^3 + cos(x)^3) / (sin(x) + cos(x)) =>
(sin(x) + cos(x)) * (sin(x)^2 - sin(x)cos(x) + cos(x)^2) / (sin(x) + cos(x)) =>
sin(x)^2 + cos(x)^2 - sin(x)cos(x)

Now, sin(x)^2 + cos(x)^2 = 1 for all values of x

1 - sin(x)cos(x)

(sin^3x+cos^3x)/sinx+cosx
(sin^2x*sinx + cos^2x*cosx)/sinx+cosx
[(1-cos^2x)*sinx + (1-sin^2x)*cosx]/sinx+cosx (sin^2x+cos^2x=1)
(sinx-cos^2xsinx+cosx-sin^2xcosx)/sinx+…
[sinx+cosx-cosx*sinx(cosx+sinx)]/sinx+c…
(sinx+cosx)/sinx+cosx - cosx*sinx(cosx+sinx)/sinx+cosx
1-sinxcosx

sin^3x+cos^3x = (sinx+cosx)(sin^2x+cos^x-sinxcosx)
by the formula of a^3+b^3
(sin^3x+cos^3x)/sinx+cosx=sin^2x+cos^x…
sin^2x+cos^x=1
(sin^x+cos^3x)/(sinx+cosx) = 1-sinxcosx