Question

(cos³A + sin³A)/(cosA + sinA) = 2- (cos³A - sin³A)/(cosA - sinA)​

Answer 1

Answer:

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

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

so cos^3A+sin^3A=(cosA+sinA)(cos^2A-cosAsinA+sin^2A)

so [cos^3A+sin^3A]/[cosA+sinA]=1-sinAcosA......................1

sin^2A+cos^2A=1

similarly

cos^A-sin^2A=(cosA-sinA)(cos^2A+sinAcosA+sin^2A)

so

[cos^3A-sin^3A]/[cosA-sinA]=1+sinAcosA.............................2

so adding both equation

1-sinAcosA+1+sinAcosA

2

LHS=RHS