Prove identity then solve

7

1. Show that \(\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } \equiv \frac { 2 } { \sin \theta \cos \theta }\).
2. Hence solve the equation \(\frac { \tan \theta } { 1 + \cos \theta } + \frac { \tan \theta } { 1 - \cos \theta } = \frac { 6 } { \tan \theta }\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

4

1. Prove that \(\frac { ( \sin \theta + \cos \theta ) ^ { 2 } - 1 } { \cos ^ { 2 } \theta } \equiv 2 \tan \theta\). \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_63_1569_333_328} .............................................................................................................................................. \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_65_1570_511_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_62_1570_603_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_685_324} \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_776_324} ...................................................................................................................................... . \includegraphics[max width=\textwidth, alt={}]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_76_1572_952_322} ........................................................................................................................................ \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_72_1570_1137_324} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_74_1572_1226_322} \includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-05_77_1575_1315_319}
2. Hence solve the equation \(\frac { ( \sin \theta + \cos \theta ) ^ { 2 } - 1 } { \cos ^ { 2 } \theta } = 5 \tan ^ { 3 } \theta\) for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\).

6

1. Prove the identity \(\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) \equiv \frac { 1 } { \tan x }\).
2. Hence solve the equation \(\left( \frac { 1 } { \cos x } - \tan x \right) \left( \frac { 1 } { \sin x } + 1 \right) = 2 \tan ^ { 2 } x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

6. (a) Show that $$\frac { \cos ^ { 2 } x - \sin ^ { 2 } x } { 1 - \sin ^ { 2 } x } \equiv 1 - \tan ^ { 2 } x , \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$ (b) Hence solve, for \(0 \leqslant x < 2 \pi\), $$\frac { \cos ^ { 2 } x - \sin ^ { 2 } x } { 1 - \sin ^ { 2 } x } + 2 = 0$$ Give your answers in terms of \(\pi\).

10

1. Show that \(\sin \left( 2 \theta + \frac { 1 } { 2 } \pi \right) = \cos 2 \theta\).
2. Hence solve the equation \(\sin 3 \theta = \cos 2 \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\).
3. Show that \(\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta\). Hence, by writing \(\cos 2 \theta - \sin 3 \theta\) in terms of \(\sin \theta\), use your answer to part (ii) to determine the solutions of \(4 x ^ { 3 } - 2 x ^ { 2 } - 3 x + 1 = 0\).

10

1. Prove that \(\cot \theta + \frac { \sin \theta } { 1 + \cos \theta } = \operatorname { cosec } \theta\).
2. Hence solve the equation \(\cot \left( \theta + \frac { \neq } { 4 } \right) + \frac { \sin \left( \theta + \frac { \neq } { 4 } \right) } { 1 + \cos \left( \theta + \frac { \neq } { 4 } \right) } = \frac { 5 } { 2 }\) for \(0 \leqslant \theta \leqslant 2 \pi\).

8

1. Show that $$\tan x = \frac { 2 t } { 1 - t ^ { 2 } } \text { for } 0 \leq t < 1 , \text { where } t = \tan \frac { 1 } { 2 } x$$ and deduce that $$\sin x = \frac { 2 t } { 1 + t ^ { 2 } }$$
2. Using the substitution \(t = \tan \frac { 1 } { 2 } x\), show that $$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { 1 + \sin x } \mathrm {~d} x = \sqrt { 3 } - 1$$

In this question you must show detailed reasoning.

1. Show that \(\cos A + \sin A \tan A = \sec A\). [3]
2. Solve the equation \(\tan 2\theta = 3 \tan \theta\) for \(0° \leqslant \theta \leqslant 180°\). [7]