AQA C3 2014 June — Question 8 12 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2014
SessionJune
Marks12
TopicReciprocal Trig & Identities

8
  1. Show that the expression \(\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x }\) can be written as \(2 \sec x\).
    [0pt] [4 marks]
  2. Hence solve the equation $$\frac { 1 - \sin x } { \cos x } + \frac { \cos x } { 1 - \sin x } = \tan ^ { 2 } x - 2$$ giving the values of \(x\) to the nearest degree in the interval \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).
    [0pt] [6 marks]
  3. Hence solve the equation $$\frac { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } { \cos \left( 2 \theta - 30 ^ { \circ } \right) } + \frac { \cos \left( 2 \theta - 30 ^ { \circ } \right) } { 1 - \sin \left( 2 \theta - 30 ^ { \circ } \right) } = \tan ^ { 2 } \left( 2 \theta - 30 ^ { \circ } \right) - 2$$ giving the values of \(\theta\) to the nearest degree in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
    [0pt] [2 marks]
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