The curves $y = ext{cos}\,x$ and $y = an x$ intersect at a point $P$ where $x$-coordinate is $eta$ - HSC - SSCE Mathematics Extension 2 - Question 7 - 2006 - Paper 1

For this part, we relate the slopes at the point of intersection:

1. Using the derivatives from part (i), we have $\text{sec}^2\beta = \frac{1}{\text{cos}^2\beta}.$
2. By substituting eta into the derived equation, we find $\text{sec}^2\alpha = 1 + \frac{\sqrt{5}}{2}.$

To prove this relationship, we can use integration by parts:

1. Set $u = \text{sec}^{n-2} t$, $dv = \text{sec}^2 t \, dt$;
2. Calculate $du$ and $v$ to apply the formula: $I_n = uv \bigg|_0^{\frac{\pi}{2}} - \int u \, dv.$
3. After evaluation, we need to rearrange the terms accordingly to achieve the desired structure. This gives the relationship between $I_n$ and $I_{n-2}$.

Utilizing our derived formula from part (ii), we substitute $n = 4$:

1. Recognize that $I_4 = \frac{\text{sec}^{4-2} α \tan α}{3} - \frac{2}{3} I_{2}$.
2. Compute $I_{2}$ similarly (using $n=2$) and obtain $I_{2}$.
3. Finish the calculation for $I_{4}$, ultimately providing the exact numeric output.

For induction, we consider the base case when n = 1:

1. Verify $x_1 = 1$ is true = $2$.
2. Assume true for $n = k$, that $x_k = \frac{2(1 + \alpha^{k})}{1 - \alpha^{k}}$.
3. Prove for $n = k + 1$ using recursion: $x_{k+1} = \frac{4 + x_k}{1 + \alpha x_k}$ and simplify to match the form.

To determine the limiting value, we examine the sequence by letting $n \to \infty$:

1. As eta = -\frac{1}{3} leads to rapid convergence properties of the geometric series.
2. Ultimately, as $\alpha^{n} \to 0$, simplifying gives the limiting result of $x_n$ as $\lim_{n \to \infty} x_n = 2.$