Question

Is cos theta +sec theta = root 3, prove that cos cube theta +sec cube theta = 0

Answer 1

$\textbf{Concept used:}$

$\text{If x+y+z=0, then }x^3+y^3+z^3=3xyz$

$\textbf{Given:}$

$cos\,\theta+sec\,\theta=\sqrt{3}$

$\implies\,cos\,\theta+sec\,\theta+(-\sqrt{3})=0$

$\text{By the given concept, we get}$

$cos^3\theta+sec^3\theta+(-\sqrt{3})^3=3\,cos\,\theta\,sec\,\theta\,(-\sqrt{3})$

$\implies\,cos^3\theta+sec^3\theta-3\sqrt{3}=-3\sqrt{3}\,cos\,\theta(\frac{1}{cos\,\theta})$

$\implies\,cos^3\theta+sec^3\theta-3\sqrt{3}=-3\sqrt{3}$

$\implies\,cos^3\theta+sec^3\theta=-3\sqrt{3}+3\sqrt{3}$

$\implies\boxed{\bf\,cos^3\theta+sec^3\theta=0}$