7. (a) Prove that
$$\frac { \cos x } { 1 - \sin x } + \frac { 1 - \sin x } { \cos x } = 2 \sec x , \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , \quad n \in \mathbb { Z }$$
(b) Hence find, for \(0 < x < \frac { \pi } { 4 }\), the exact solution of
$$\frac { \cos x } { 1 - \sin x } + \frac { 1 - \sin x } { \cos x } = 8 \sin x$$
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(a) \(\frac{\cos x}{1 - \sin x} + \frac{1 - \sin x}{\cos x} = \frac{\cos x \cos x + (1 - \sin x)(1 - \sin x)}{(1 - \sin x)\cos x}\) M1, A1
\(= \frac{\cos^2 x + \sin^2 x + 1 - 2\sin x}{(1 - \sin x)\cos x}\) M1
\(= \frac{2 - 2\sin x}{(1 - \sin x)\cos x}\) A1
\(= \frac{2(1 - \sin x)}{(1 - \sin x)\cos x}\) \(= \frac{2}{\cos x} = 2\sec x\) (b) \(\frac{\cos x}{1 - \sin x} + \frac{1 - \sin x}{\cos x} = 8\sin x\) M1
\(2\sec x = 8\sin x\) M1
\(1 = 4\sin x \cos x\) or \(a\sin x \cos x = b\) M1, A1
\(\sin 2x = \frac{1}{2}\) or \(\sin 2x = 2\sin x \cos x\) \(x = \frac{1}{2}\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{12}\)
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(a) $\frac{\cos x}{1 - \sin x} + \frac{1 - \sin x}{\cos x} = \frac{\cos x \cos x + (1 - \sin x)(1 - \sin x)}{(1 - \sin x)\cos x}$ | M1, A1
$= \frac{\cos^2 x + \sin^2 x + 1 - 2\sin x}{(1 - \sin x)\cos x}$ | M1
$= \frac{2 - 2\sin x}{(1 - \sin x)\cos x}$ | A1
$= \frac{2(1 - \sin x)}{(1 - \sin x)\cos x}$ |
$= \frac{2}{\cos x} = 2\sec x$ |
(b) $\frac{\cos x}{1 - \sin x} + \frac{1 - \sin x}{\cos x} = 8\sin x$ | M1
$2\sec x = 8\sin x$ | M1
$1 = 4\sin x \cos x$ or $a\sin x \cos x = b$ | M1, A1
$\sin 2x = \frac{1}{2}$ or $\sin 2x = 2\sin x \cos x$ |
$x = \frac{1}{2}\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{12}$ |
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7. (a) Prove that
$$\frac { \cos x } { 1 - \sin x } + \frac { 1 - \sin x } { \cos x } = 2 \sec x , \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , \quad n \in \mathbb { Z }$$
(b) Hence find, for $0 < x < \frac { \pi } { 4 }$, the exact solution of
$$\frac { \cos x } { 1 - \sin x } + \frac { 1 - \sin x } { \cos x } = 8 \sin x$$
\hfill \mbox{\textit{Edexcel C3 2013 Q7 [8]}}