Problems

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Solution: \(\displaystyle (1-x^6)^{-2} = \sum_{n=0}^{\infty} (n+1)x^{6n}\)

1. \(\,\) \begin{align*} && f(x) &= (1-x^6)^{-2}(1-x^3)^{-1} \\ &&&= \left ( \sum_{n=0}^{\infty} (n+1)x^{6n} \right) \left ( \sum_{n=0}^{\infty} x^{3n} \right) \\ [x^{24}]: && c_{24} &= 1 + 2+ 3+4+5 = 15 \end{align*} Clearly \(n\) must be a multiple of \(3\). If \(n = 6k\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) If \(n = 6k+3\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) the same way, we just must always get one extra \(x^3\) term from the second expansion.
2. We can obtain \(x^{24}\) from the product of \((1-x^6)^{-2}(1-x^3)^{-1}\) and \((1-x)^{-1}\) in the following ways: \begin{array}{c|c|c} (1-x^6)^{-2}(1-x^3)^{-1} & (1-x)^{-1} & \text{product} \\ \hline 15x^{24} & x^0 & 15x^{24} \\ 10x^{21} & x^3 & 10x^{24} \\ 10x^{18} & x^6 & 10x^{24} \\ 6x^{15} & x^9 & 6x^{24} \\ 6x^{12} & x^{12} & 6x^{24} \\ 3x^{9} & x^{15} & 3x^{24} \\ 3x^{6} & x^{18} & 3x^{24} \\ x^{3} & x^{21} & x^{24} \\ x^{0} & x^{24}& x^{24} \end{array} So the total is \(55\). Similarly for \(25\) we can only obtain this in the same ways but also taking an extra power of \(x\) from the geometric series, ie \(55\) For \(66\) we obtain by similar reasoning that it is: \(\frac{13\cdot12}{2} + 2 \left (1 + 3 + \cdots + \frac{13 \cdot 12}{2} \right) = \frac{13\cdot12}{2} + 2 \binom{14}{3} = \frac{13 \cdot 12}2 ( 1 + \frac{30}{3}) = 11 \cdot 6 \cdot 13 = 858\)

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Solution: If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), \(\p(\q(x))\) has degree \(mn\).

1. Suppose \(\p(\p(\p(x)))- 3 \p(x)= -2x\), and suppose \(p(x)\) has degree \(n = \geq 2\), then \(\p(\p(\p(x)))\) has degree \(n^3\) and so the left hand side has degree higher than \(1\) and the right hand side is degree \(1\). Therefore \(\p(x)\) is degree \(1\) or \(0\). If \(p(x) = c\) then \(c^3-3c = -2x\) but the LHS doesn't depend on \(x\) which is also a contradiction. Therefore \(\p(x)\) is degree \(1\). Suppose \(\p(x) = ax+b\) then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
2. Suppose \(2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4\) and let \(\deg \p(x) = n\), then LHS has degree \(\max(n^2,2n,n)\) and the right hand side has degree \(4\). Therefore \(\p(x)\) must have degree \(2\). Let \(\p(x) = ax^2 + bx + c\), then, considering the coefficient of \(x^4\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we will have \(2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12\). Consider the coefficient of \(x^3\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we have \(4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0\) Since \(a = -1, \frac12\) this means \(b = 0\). Consider the constant coefficient in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) (using \(b = 0\)). \(2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0\). Therefore \(c = 0\) or \(a = -1, c = 3, a = \frac12, c = \frac34\), so our possible polynomials are: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)

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Solution: \begin{align*} && t &= x + \sqrt{1+x^2} \\ &&\frac1t &= \frac{1}{x+\sqrt{1+x^2}} \\ &&&= \frac{\sqrt{1+x^2}-x}{1+x^2-1} \\ &&&= \sqrt{1+x^2}-x \\ \Rightarrow && x &=\frac12 \left ( t - \frac1t\right) \\ \Rightarrow && \d x &=\frac12 \left (1 + \frac1{t^2} \right)\d t \\ \\ \Rightarrow && \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x &= \int_{t=1}^{t = \infty}f(t) \frac12\left (1 + \frac1{t^2} \right)\d t \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right)f(t) \d t \end{align*} \begin{align*} && I &= \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \d x \\ &&&= \int_0^\infty \frac1 {(x+\sqrt{x^2+1})^2} \d x \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right) \frac{1}{t^2} \d t \\ &&&= \frac12 \left [-\frac1t-\frac13\frac1{t^3} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac43 = \frac23 \end{align*} \begin{align*} && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&x &= \tan \theta\\ && \d x &= \sec^2 \theta = (1+x^2) \d \theta\\ && \tan\theta &= \frac{s}{\sqrt{1-s^2}}\\ \Rightarrow && \tan^2 \theta &= \frac{s^2}{1-s^2} \\ \Rightarrow && \sin \theta &= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\ && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&&= \int_0^{\frac12 \pi} \frac{1}{\left (1+ \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \right )^3} \d \theta \\ &&&= \int_{x=0}^{x=\infty} \frac{1}{\left(1 + \frac{x}{\sqrt{1+x^2}} \right)^3} \frac{1}{1+x^2} \d x \\ &&&= \int_0^{\infty} \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2}+x)^3} \d x \\ &&J_a &= \int_0^{\infty} \frac{\sqrt{1+x^2}+x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \frac23 \\ &&J_b &= \int_0^{\infty} \frac{\sqrt{1+x^2}-x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \int_0^{\infty} \frac{1}{(\sqrt{1+x^2}+x)^4} \d x\\ &&&= \frac12\int_1^{\infty} \left (1 +\frac1{t^2} \right)\frac{1}{t^4} \d t \\ &&&= \frac12 \left [-\frac13 t^{-3}-\frac15t^{-5} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac8{15} = \frac4{15} \\ \Rightarrow && J &= \frac12(J_a+J_b) = \frac7{15} \end{align*}

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Solution:

1. Since \(k \geq 1\) we have \(n^{k+1} > n^k\) and \((k+1) > k\), therefore \((k+1)n^{k+1} >kn^k \Rightarrow \frac{1}{(k+1)n^{k+1}} < \frac{1}{kn^k}\) \begin{align*} && \ln \left ( 1 + \frac1n \right) &= \frac1n -\frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \cdots \\ &&&= \frac1n - \underbrace{\left (\frac{1}{2n^2}-\frac{1}{3n^3} \right)}_{>0}- \underbrace{\left (\frac{1}{4n^4}-\frac{1}{5n^5} \right)}_{>0} - \cdot \\ &&&< \frac1n \\ \\ \Rightarrow && n \ln \left ( 1 + \frac1n \right) &< 1 \\ \Rightarrow && \ln \left ( \left ( 1 + \frac1n \right)^n \right) &< 1 \\ \Rightarrow && \left ( 1 + \frac1n \right)^n &< e \end{align*}
2. \(\,\) \begin{align*} &&\ln \left(\frac{2y+1}{2y-1}\right) &= \ln \left (1 + \frac{1}{2y} \right)-\ln \left (1 - \frac{1}{2y} \right) \\ &&&= \frac{1}{2y} - \frac{1}{2(2y)^2} + \frac{1}{3(2y)^3} - \cdots - \left (-\frac{1}{2y} - \frac{1}{2(2y)^2} - \frac{1}{3(2y)^3} - \cdots \right) \\ &&&= \frac{1}{y} + \frac{2}{3(2y)^3} + \frac{2}{5(2y)^5} \\ &&&= \sum_{r=1}^{\infty} \frac{2}{(2r-1)(2y)^{2r-1}} \\ &&&> \frac1y \\ \\ \Rightarrow && \ln \left (1 + \frac{1}{y-\frac12} \right) &> \frac{1}{y} \\\Rightarrow && \ln \left (1 + \frac{1}{n} \right) &> \frac{1}{n+\frac12} \\ \Rightarrow &&(n+\tfrac12) \ln \left (1 + \frac{1}{n} \right) &> 1\\ \Rightarrow && \ln \left ( \left (1 + \frac{1}{n} \right)^{n+\tfrac12} \right) &> 1\\ \Rightarrow && \left (1 + \frac{1}{n} \right)^{n+\tfrac12} & > e \end{align*}

\item Since \(\left (1 + \frac1n \right)^n\) is both bounded above, and increasing, it must tend to some limit \(L\). \begin{align*} && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n+\frac12} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \lim_{n \to \infty} \sqrt{1 + \frac1n} \\ \Rightarrow && \lim_{n \to \infty} \left (1 + \frac1n \right)^n && \leq e &\leq & \lim_{n \to \infty} \left (1 + \frac1n \right)^{n} \\ \end{align*} And therefore equality must hold.

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Solution:

1. \(\,\)
   

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2. \(\,\) \begin{align*} && \frac{\d}{\d x} \left ( \frac{1}{g(x)} \right) &= \frac{\d }{\d x} \left ( \big( (x-a)^2-1 \big) \big( (x-b)^2 -1\big)\right) \\ &&&= ((x-a)^2-1)(2(x-b))+((x-b)^2-1)(2(x-a)) \\ &&&= 2(2x-a-b)(x^2-(a+b)x+ab-1) \\ \Rightarrow && x &= \frac{a+b}{2}, \frac{a+b \pm \sqrt{(a+b)^2-4ab+4}}{2} \\ &&&= \frac{a+b}{2}, \frac{a+b \pm \sqrt{(a-b)^2+4}}{2} \end{align*} If \(b > a+2\):
   

   + - ↺
   If \(b = a+2\):
   

   + - ↺

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Solution:

\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)

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Solution:

\begin{align*} && \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\ &&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 - \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ && 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\ \Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\ &&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\ && 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\ \Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\ &&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\ \Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)} \end{align*} as required. Since \(\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}\) we must have, \begin{align*} && \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} + \frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\ &&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\ &&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\ &&&= 3 \end{align*}

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Solution: \begin{align*} && \beta - \alpha &> q \\ \Rightarrow &&(\beta - \alpha)^2 &> q^2 \\ \Rightarrow && \beta^2 +\alpha^2 - 2\beta \alpha &> q^2 \\ \Rightarrow && \alpha^2+\beta^2-q^2 -2 \beta \alpha &> 0 \\ \Rightarrow && \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} - 2 &> 0 \end{align*} \begin{align*} && u_n(u_n+u_{n+2}) &= u_n \cdot \left (u_n + \frac {u_{n+1}^2 -q^2}{u_{n}}\right) \\ &&&= u_n^2 + u_{n+1}^2-q^2 \\ &&&= u_n^2 + u_{n+1}^2 - (u_n^2-u_{n-1}u_{n+1}) \\ &&&= u_{n+1}^2 + u_{n+1}u_{n-1} \\ &&&= u_{n+1}(u_{n-1}+u_{n+1}) \\ \\ && u_{n+1}-pu_n+u_{n-1} &= -pu_n+\frac{u_{n}(u_{n-2}+u_n)}{u_{n-1}} \\ &&&= \frac{u_n(u_{n}-pu_{n-1}+u_{n-2})}{u_{n-1}} \end{align*} Therefore if \(u_2 -pu_1 + u_0 = 0\) it is always zero, ie if \begin{align*} && u_2 &= p\beta - \alpha \\ && u_2 &= \frac{\beta^2-q^2}{\alpha} \\ \Rightarrow && \frac{\beta^2-q^2}{\alpha} &= p\beta - \alpha \\ \Rightarrow && p &= \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} \end{align*} If \(\beta > \alpha + q\) we must have that \(p > 2\), and so \(u_{n+1}-u_n = (p-1)u_n - u_{n-1} > u_n-u_{n-1} > 0\), therefore the sequence is strictly increasing. If \(\beta = \alpha + q\) the sequence follows \(u_{n+1} - 2u_n + u_{n-1} =0\) and so \(u_{n+1}-u_n = u_n - u_{n-1}\) for all \(n\) (which is still increasing - it's an arithmetic progression with common difference \(\beta - \alpha\)).

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Solution: \begin{align*} && s &= ut \\ \Rightarrow && a &= u \cos \alpha t\\ \Rightarrow && t &= \frac{a}{u \cos \alpha}\\ && s &= ut+ \frac12at^2 \\ \Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\ &&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\ \Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\ \Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \end{align*} \begin{align*} && s &= ut + \frac12a t^2 \\ \Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\ \Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && s &= ut \\ \Rightarrow && b &> u \cos \alpha t \\ \Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\ \Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\ \end{align*} \begin{align*} \Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\ \Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\ &&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\ &&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\ \Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\ \Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\ \Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\ \Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)} \end{align*}

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Solution:

Let \(M\) be the midpoint of \(AB\), then \begin{align*} \overset{\curvearrowright}{M}: && R_1 \sin \phi-\mu R_1 \cos \phi &= R_2 \sin \phi+\mu R_2 \cos \phi \\ \Rightarrow && (R_1-R_2) \tan \phi &= \mu(R_1+R_2) \end{align*} As required. \begin{align*} && \cos \phi = \frac{a}{r} &,\,\, \sin \phi = \frac{\sqrt{r^2-a^2}}{r} \\ \text{N2}(\rightarrow): && R_1\cos(\phi + \theta)+\mu R_1 \sin(\phi + \theta) &= R_2 \cos(\theta - \phi) + \mu R_2 \sin(\theta - \phi) \\ \Rightarrow && R_1(\cos \theta \cos \phi - \sin \theta \sin \phi)+ \mu R_1 (\sin \theta \cos \phi + \cos \theta \sin \phi) &= R_2 (\cos\theta \cos \phi + \sin \theta \sin \phi)+ \mu R_2 (\sin \theta \cos \phi - \cos \theta \sin \phi) \\ && R_1 (1 - \tan \theta \tan \phi)+\mu R_1 (\tan \theta + \tan \phi) &= R_2(1 + \tan \theta \tan \phi) +\mu R_2 (\tan \theta - \tan \phi) \\ && 0 &= (R_1-R_2)(1+\mu \tan \theta)+(R_1+R_2)(-\tan \theta \tan\phi+\mu \tan \phi) \\ \Rightarrow && \frac{R_1+R_2}{R_1-R_2} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \frac{\tan \phi}{\mu} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \tan^2 \phi &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \frac{r^2-a^2}{a^2} &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \tan \theta (r^2-a^2-a^2\mu^2) &= \mu a^2+\mu(r^2-a^2) \\ \Rightarrow && \tan \theta &= \frac{\mu r^2}{r^2-(1+\mu^2)a^2} \end{align*} Since \(\mu r^2 > 0\) we must also have \(r^2 > a^2(1+\mu^2)\) ie \(\\sec^2 \phi > 1 + \mu^2 = \sec^2 \lambda\) and the result follows.