Question

Prove that:
${\displaystyle \frac{{\cos }^3\theta +{\sin }^3\theta }{\cos \theta +\sin \theta }+\frac{{\cos }^3\theta -{\sin }^3\theta }{\cos \theta -\sin \theta }=2}$

Answer 1

LHS = ${\displaystyle \frac{{\cos }^3\theta +{\sin }^3\theta }{\cos \theta +\sin \theta }+\frac{{\cos }^3\theta -{\sin }^3\theta }{\cos \theta -\sin \theta }}$

= ${\displaystyle \frac{(\cos \theta +\sin \theta )({\cos }^2\theta +{\sin }^2\theta -\cos \theta \sin \theta )}{\cos \theta +\sin \theta }+\frac{(\cos \theta -\sin \theta )({\cos }^2\theta +{\sin }^2\theta -\cos \theta \sin \theta )}{\cos \theta -\sin \theta }}$

= 1 - sin θ cos θ + 1 + sin θ cos θ

= 2 

= RHS

Hence proved.