Let $\alpha $ and $\beta $ be the roots of the quadratic equation ${x^2}\,\sin \,\theta - x\,\left( {\sin \,\theta \cos \,\,\theta + 1} \right) + \cos \,\theta = 0\,\left( {0 < \theta < {{45}^o}} \right)$ , and $\alpha < \beta $. Then $\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + \frac{{{{\left( { - 1} \right)}^n}}}{{{\beta ^n}}}} \right)} $ is equal to
$\frac{1}{{1 - \cos \,\theta }} - \frac{1}{{1 + \sin \,\theta \,}}$
$\frac{1}{{1 + \cos \,\theta }} + \frac{1}{{1 - \sin \,\theta \,}}$
$\frac{1}{{1 - \cos \,\theta }} + \frac{1}{{1 + \sin \,\theta \,}}$
$\frac{1}{{1 + \cos \,\theta }} - \frac{1}{{1 - \sin \,\theta \,}}$
The sum of the cubes of all the roots of the equation $x^{4}-3 x^{3}-2 x^{2}+3 x+1=10$ is
If ${x^2} + px + 1$ is a factor of the expression $a{x^3} + bx + c$, then
If $x$ is real and satisfies $x + 2 > \sqrt {x + 4} ,$ then
If ${x^2} + 2ax + 10 - 3a > 0$ for all $x \in R$, then
Let $x, y, z$ be non-zero real numbers such that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=7$ and $\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=9$, then $\frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}-3$ is equal to