Question

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

Answer 1

L.H.S$=\frac{{\cos }^3\theta +{\sin }^3\theta }{\cos \theta +\sin \theta }+\frac{{\cos }^3\theta -{\sin }^3\theta }{\cos \theta -\sin \theta }$

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

$={\cos }^2\theta +{\sin }^2\theta -\cos \theta \sin \theta \theta +{\cos }^2\theta +{\sin }^2\theta +\cos \theta \sin \theta \theta$

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

$=2=$R.H.S since ${\cos }^2\theta +{\sin }^2\theta =1$

Hence proved.