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

3 Answers
Mar 29, 2018

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)

Mar 29, 2018

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

Mar 29, 2018

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)