Question

Find the value of

$\frac{{\sin }^3\theta +{\cos }^3\theta }{\sin \theta +\cos \theta }+\frac{({\cos }^3\theta -{\sin }^3\theta )}{\cos \theta -\sin \theta }$

• A. $2$
• B. $1$
• C. $0$
• D. None of these

Answer 1

The correct option is A $2$

To find the value of $\frac{{\sin }^3\theta +{\cos }^3\theta }{\sin \theta +\cos \theta }+\frac{{\cos }^3\theta -{\sin }^3\theta }{\cos \theta -\sin \theta }$

Using the formulae,

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

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

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

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

$=(1-\sin \theta \cos \theta )+(1+\sin \theta \cos \theta )$

$=1-\sin \theta \cos \theta +1+\sin \theta \cos \theta$

$=2$

Hence, the required answer is $2$.