MCQ
${\sec ^2}({\tan ^{ - 1}}2) + {\rm{cose}}{{\rm{c}}^2}({\cot ^{ - 1}}3) = $
- A$5$
- B$13$
- ✓$15$
- D$6$
and ${\cot ^{ - 1}}3 = \beta \Rightarrow \cot \beta = 3$
${\sec ^2}({\tan ^{ - 1}}2) + {\rm{cose}}{{\rm{c}}^{\rm{2}}}({\cot ^{ - 1}}3)$
$= {\sec ^2}\alpha + {\rm{cose}}{{\rm{c}}^{\rm{2}}}\alpha $
$=1 + {\tan ^2}\alpha + 1 + {\cot ^2}\alpha $
$= 2 + {(2)^2} + {(3)^2} = 15$.
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$\text{k}\in(2,\infty)$
$\text{k}\in(-\infty,2)$
$\text{k}\in(4,\infty)$
$\text{k}\in(-\infty,4)$
| X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is: