Let $$ - {\pi \over 6} < \theta < - {\pi \over {12}}.$$ Suppose $${\alpha _1}$$ and $${\beta_1}$$ are the roots of the equation $${x^2} - 2x\sec \theta + 1 = 0$$ and $${\alpha _2}$$ and $${\beta _2}$$ are the roots of the equation $${x^2} + 2x\,\tan \theta - 1 = 0.$$ $$If\,{\alpha _1} > {\beta _1}$$ and $${\alpha _2} > {\beta _2},$$ then $${\alpha _1} + {\beta _2}$$ equals

The quadratic equation $$p(x)$$ $$ = 0$$ with real coefficients has purely imaginary roots. Then the equation $$p(p(x))=0$$ has

Let $$\alpha $$ and $$\beta $$ be the roots of $${x^2} - 6x - 2 = 0,$$ with $$\alpha > \beta .$$ If $${a_n} = {\alpha ^n} - {\beta ^n}$$ for $$\,n \ge 1$$ then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is