A box contains 9 tickets numbered 1 to 9 both inclusive. If 3 tickets are drawn from the box one at a time, then the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

$\lim _\limits{x \rightarrow 3} \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$ is

The statement pattern $[(p \rightarrow q) \wedge \sim q] \rightarrow r$ is a tautology when $r$ is equivalent to

If $3 \sin \alpha=5 \sin \beta$, then $\tan \left(\frac{\alpha+\beta}{2}\right)+\tan \left(\frac{\alpha-\beta}{2}\right)=$