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(a) \(x + \frac{1}{x} = \sec\theta + \tan\theta + \frac{1}{\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}\)M1
\(= \frac{\sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta + 1}{\sec\theta + \tan\theta} = \frac{2\sec^2\theta + 2\sec\theta\tan\theta}{\sec\theta + \tan\theta}\) M1 A1
\(= \frac{2\sec\theta(\sec\theta + \tan\theta)}{\sec\theta + \tan\theta} = 2\sec\theta\) M1 A1
(b) \(\frac{x^2+1}{x} = \frac{2}{\cos\theta} \Rightarrow \cos\theta = \frac{2x}{x^2+1}\)M1
\(\frac{y^2+1}{y} = \frac{2}{\sin\theta} \Rightarrow \sin\theta = \frac{2y}{y^2+1}\) \(\therefore \frac{4x^2}{(x^2+1)^2} + \frac{4y^2}{(y^2+1)^2} = 1\) M1 A1
(c) \(\frac{dv}{d\theta} = \sec\theta\tan\theta + \sec^2\theta\)M1
\(= \sec\theta(\tan\theta + \sec\theta) = \frac{x^2+1}{2x} \cdot x = \frac{1}{2}(x^2+1)\) M1 A1
(d) \(\frac{dy}{d\theta} = -\cosec\theta\cot\theta - \cosec^2\theta\)M1
\(= -\cosec\theta(\cot\theta + \cosec\theta) = -\frac{y^2+1}{2y} \cdot y = -\frac{1}{2}(y^2+1)\) A1
\(\therefore \frac{dy}{dx} = \frac{-y^2+1}{x^2+1}\) M1 A1
(15)
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**(a)** $x + \frac{1}{x} = \sec\theta + \tan\theta + \frac{1}{\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}$ | M1 |
$= \frac{\sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta + 1}{\sec\theta + \tan\theta} = \frac{2\sec^2\theta + 2\sec\theta\tan\theta}{\sec\theta + \tan\theta}$ | M1 A1 |
$= \frac{2\sec\theta(\sec\theta + \tan\theta)}{\sec\theta + \tan\theta} = 2\sec\theta$ | M1 A1 |
**(b)** $\frac{x^2+1}{x} = \frac{2}{\cos\theta} \Rightarrow \cos\theta = \frac{2x}{x^2+1}$ | M1 |
$\frac{y^2+1}{y} = \frac{2}{\sin\theta} \Rightarrow \sin\theta = \frac{2y}{y^2+1}$ | |
$\therefore \frac{4x^2}{(x^2+1)^2} + \frac{4y^2}{(y^2+1)^2} = 1$ | M1 A1 |
**(c)** $\frac{dv}{d\theta} = \sec\theta\tan\theta + \sec^2\theta$ | M1 |
$= \sec\theta(\tan\theta + \sec\theta) = \frac{x^2+1}{2x} \cdot x = \frac{1}{2}(x^2+1)$ | M1 A1 |
**(d)** $\frac{dy}{d\theta} = -\cosec\theta\cot\theta - \cosec^2\theta$ | M1 |
$= -\cosec\theta(\cot\theta + \cosec\theta) = -\frac{y^2+1}{2y} \cdot y = -\frac{1}{2}(y^2+1)$ | A1 |
$\therefore \frac{dy}{dx} = \frac{-y^2+1}{x^2+1}$ | M1 A1 | (15)
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**Total** | (75)
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7 \\
0 \\
- 3
\end{array} \right) + \lambda \left( \begin{array} { c }
5 \\
4 \\
- 2
\end{array} \right) \\
& \mathbf { r } = \left( \begin{array} { l }
a \\
6 \\
3
\end{array} \right) + \mu \left( \begin{array} { c }
- 5 \\
\hfill \mbox{\textit{Edexcel C4 Q7 [15]}}