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)#