Question

Sin³A+cos³A/sinA+cosa+sinAcosA
Evaluate

Answer 1

[tex] \frac{sin^3A+cos^3A}{sinAcosA} +sinAcosA \\ \frac{(sinA+cosA)(sin^2A+cos^2A-sinAcosA)}{sinAcosA}+sinAcosA \\ sin^2A+cos^2A-sinAcosA+sinAcosA \\ =1 [/tex]

Answer 2

Given that $\frac{Sin^3A + Cos^3A}{SinA+CosA+SinACosA}$

But it should be  \frac{Sin^3A + Cos^3A}{SinA+CosA-SinACosA} 

We know that,

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

$\frac{(SinA+CosA)(Cos^2A+Sin^2A - CosASinA)}{(Cos^2A+Sin^2A - CosASinA)}$

SinA + CosA