Question

Prove:

$\frac{{cos}^{3}A+{sin}^{3}A}{cosA+sinA}+\frac{{cos}^{3}A-{sin}^{3}A}{cosA-sinA}=2$

Answer 1

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

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

so $co{s}^{3}A+si{n}^{3}A=(cosA+sinA)(co{s}^{2}A-cosAsinA+si{n}^{2}A)$

so$\frac{[co{s}^{3}A+si{n}^{3}A]}{[cosA+sinA]}=1-sinAcosA⟹\left(1\right)$

$si{n}^{2}A+co{s}^{2}A=1$

similarly $co{s}^A-si{n}^{2}A=(cosA-sinA)(co{s}^{2}A+sinAcosA+si{n}^{2}A)$

so $\frac{[co{s}^{3}A-si{n}^{3}A]}{[cosA-sinA]}=1+sinAcosA⟹\left(2\right)$

so adding both equation (1) and (2) $1-sinAcosA+1+sinAcosA=2$

Hence Proved.