Question

How do you prove `(cosx - sinx)/(cos^3x - sin^3x) = 1/(tanxcos^2x + 1)`?

Answer 1

Verified below...

Explanation:

Given:
`(cosx-sinx)/(cos^3x-sin^3x)= 1/(tanxcos^2x+1)`

When working with the denominator: `(a^3-b^3)= (a-b)(a^2+ba+b^2)`

`(cosx-sinx)/((cosx-sinx)(cos^2x+cosxsinx+sin^2x))= 1/(tanxcos^2x+1)`

`cancel(cosx-sinx)/(cancel(cosx-sinx)(cos^2x+cosxsinx+sin^2x))= 1/(tanxcos^2x+1)`

`1/(cos^2x+cosxsinx+sin^2x)= 1/(tanxcos^2x+1)`

Remember: `sin^2x+cos^2x=1`

`1/(1+cosxsinx)= 1/(tanxcos^2x+1)`

Remember `tanx= sinx/cosx` therefore: `sinx=tanx*cosx`:
`1/(1+cosx(tanx*cosx))= 1/(tanxcos^2x+1)`

`1/(tanxcos^2x+1)= 1/(tanxcos^2x+1)`

Answer 2

Proved below.
Please see the explanation.

Explanation:

lhs=

`(cosx-sinx)/(cos^3x-sin^3x)`

Now,
Let
`a=cosx, b=sinx`

`(cosx-sinx)/(cos^3x-sin^3x)=(a-b)/(a^3-b^3)`

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

`(a-b)/(a^3-b^3)=((a-b))/((a-b)(a^2+b^2+ab))`

Cancelling a-b

`(a-b)/(a^3-b^3)=1/(a^2+b^2+ab)`

`a^2+b^2+ab=(cosx)^2+(sinx)^2+(cosx)(sinx)`

`=cos^2x+sin^2x+cosx/cosxcosxsinx`

`cos^2x+sin^2x=1`

`cosx/cosxcosxsinx=cosxcosxsinx/cosx`

`sinx/cosx=tanx`

`cosx/cosxcosxsinx=cos^2xtanx`

`(cosx)^2+(sinx)^2+(cosx)(sinx)=1+cos^2xtanx`

`(a-b)/(a^3-b^3)=1/(a^2+b^2+ab)`

`(cosx-sinx)/(cos^3x-sin^3x)=1/(1+cos^2xtanx)`

`=rhs`

Answer 3

factor cos `cos^3x-sin^3x`=`(cosx-sinx)(cos^2x+cosx*sinx+sin^2x)`

Explanation:

`(cosx-sinx)/(cos^3x-sin^3x)=(cosx-sinx)/((cosx-sinx)(cos^2x+cosxsinx+sin^2x)`
[cancelling the `(cosx-sinx)` both in numerator and denominator]

`=1/(cos^2x+cosxsinx+sin^2x)`
[using trignometric identities `cos^2x+sin^2x=1`]
`=1/(1+cosxsinx)`

multiplying and dividing by `cosx`

`=1/(1+cosxsinx(cosx/cosx))`
`=1/(1+tanxcos^2x)`