The diagram shows the graph $y = ext{ln} x.$ By comparing relevant areas in the diagram, or otherwise, show that \[ ext{ln} t > \frac{t - 1}{t + 1} \text{ for } t > 1 - HSC - SSCE Mathematics Extension 2 - Question 14 - 2013 - Paper 1

To prove this by induction:

Base case: For $n = 2$, we have: [ |z_2| = |1 + i| = \sqrt{2} = \sqrt{2}. ]

Inductive step: Assume it holds for $n = k$, i.e., $|z_k| = \sqrt{k}$. Then for $n = k + 1$: [ z_{k+1} = z_k \left( \frac{i}{|z_k|} \right). ]

Calculating $|z_{k+1}|$: [ |z_{k+1}| = |z_k| \left( \frac{1}{|z_k|} \right) = |z_k| = \sqrt{k}, ] This leads to: [ |z_{k+1}| = \sqrt{k + 1}. ] Thus, the induction hypothesis holds.

Using the identity for secant: [ ext{sec}^2 \theta = 1 + \tan^2 \theta ] can be represented by the binomial expansion. For positive integer $n$, [ (1 + u)^n = \sum_{k=0}^{n} \binom{n}{k} u^k, ] where we let $u = \tan^2 \theta$, leading to: [ ext{sec}^2 \theta = \sum_{k=0}^{n} \binom{n}{k} \tan^{2k} \theta. ]

Using integration and the previous identities: [ \int \text{sec}^6 \theta , d\theta = \int \text{sec}^4 \theta \text{sec}^2 \theta , d\theta. ]

Using integration by parts or known integrals, we compute: [ = \frac{1}{3} \text{sec}^3 \theta \tan \theta + C. ]

To establish similarity, we can use the Angle-Angle (AA) criterion. The angle $ext{A}$ is common to both triangles, and: [ \angle ABC = \angle AED ] proves that triangles $ABC$ and $AED$ are similar.

To prove that quadrilateral $BCED$ is cyclic, we show that angles $\angle BEC$ and $\angle BDC$ subtend the same arc. Therefore, $\angle BEC + \angle BDC = 180^{\circ}$. Hence, $BCED$ is a cyclic quadrilateral.

Using the distance formula in triangle $ABC$: [ CD = \sqrt{(AD)^2 + (AE)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. ]

Using coordinate geometry, we can find the circumradius $R$ between points: [ R = \frac{abc}{4K}, ] where $a$, $b$, and $c$ are the sides of triangle $BCE$ and $K$ is the area. After computing, we find the radius to be: [ R = \frac{5\sqrt{21}}{4}. ]