3d iii)
})
})
})
})
Thus, parallel (corresponding angles are equal).
You could have stopped knowing the alternate segment theorem, and use part (ii).
5d (ii) I'll skip the first step, as I assume you know how to do it.
\theta = -(k + 1) + cot\theta tan(k + 1)\theta ))
\theta + tan(k+1)\theta tan(k+2)\theta)
 + cot\theta tan(k+1)\theta + tan(k+1)\theta tan(k+2)\theta \text{ (by assumption)})
 + cot\theta tan(k+1)\theta + cot\theta (tan(k+2)\theta -<br /> tan(k+1)\theta ) - 1 \text{ (from i)})
 + cot\theta (tan(k+1)\theta) + tan(k+2)\theta - tan(k+1)\theta ))
 + cot\theta tan(k+2)\theta = RHS)
The part that's not showing in my preview is basically the substitution from part i, where n is equal to (k+1). When we substitute the RHS of part i, we can factorise and it should come out nicely.
6b)
(i) P(at least three of its members not complete) = P(3 not complete) + P(4 not complete)
 } = \binom{4}{3}q^{3}p + \binom{4}{4}q^{4}p^{0} = 4pq^{3} + q^{4})
(ii) To score a point, at least half will complete. Thus, we need at least 2 people to complete it, meaning we need 0 or 1 to not pass.
 } = 1 - \text{ P(1 complete) - P(0 complete)})
q^{3} - q^{4})
(iii) Again, probability that a two-member team will score is when only zero fail.
} = 1 - \text{ (P(0 complete)})
(iv)
 < 0)
 < 0)
 - (q - 1)] < 0)
7b)
Using the arc length-radius-central angle property,
}{4\theta^{2}})
\theta^{2} - (\theta - sin\theta cos\theta)2\theta}{\theta^{4}}\right))
\theta) -2\theta + sin2\theta)}{\theta^{4}}\right))
)
)
 + 2sin\theta cos\theta}{\theta^{3}}\right))
)
}{2\theta^{3}})
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