Question

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

Answer 1

Proof :

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

we use two identities,

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

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

using the above identities,

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

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

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

$=2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$

Hence, proved.