Substitution t equals tan

5. (a) Use the substitution \(t = \tan x\) to show that the equation $$12 \tan 2 x + 5 \cot x \sec ^ { 2 } x = 0$$ can be written in the form $$5 t ^ { 4 } - 24 t ^ { 2 } - 5 = 0$$ (b) Hence solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation $$12 \tan 2 x + 5 \cot x \sec ^ { 2 } x = 0$$ Show each stage of your working and give your answers to one decimal place.

7. (a) Use the substitution \(t = \tan x\) to show that the equation $$4 \tan 2 x - 3 \cot x \sec ^ { 2 } x = 0$$ can be written in the form $$3 t ^ { 4 } + 8 t ^ { 2 } - 3 = 0$$ (b) Hence solve, for \(0 \leqslant x < 2 \pi\), $$4 \tan 2 x - 3 \cot x \sec ^ { 2 } x = 0$$ Give each answer in terms of \(\pi\). You must make your method clear.

7

1. Write down the formula for \(\tan 2 x\) in terms of \(\tan x\).
2. By letting \(\tan x = t\), show that the equation $$4 \tan 2 x + 3 \cot x \sec ^ { 2 } x = 0$$ becomes $$3 t ^ { 4 } - 8 t ^ { 2 } - 3 = 0$$
3. Hence find all the solutions of the equation $$4 \tan 2 x + 3 \cot x \sec ^ { 2 } x = 0$$ which lie in the interval \(0 \leqslant x \leqslant 2 \pi\).

1. (a) Use \(t = \tan \frac { \theta } { 2 }\) to show that, where both sides are defined

$$\frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta } \equiv \frac { 5 t + 2 } { 2 t + 5 }$$ (b) Hence, again using \(t = \tan \frac { \theta } { 2 }\), prove that, where both sides are defined $$\frac { 20 + 21 \tan \theta } { 29 + 21 \sec \theta } \equiv \frac { 29 - 21 \sec \theta } { 20 - 21 \tan \theta }$$

1. (a) Use the substitution \(\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)\) to show that

$$\sec x - \tan x \equiv \frac { 1 - t } { 1 + t } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$ (b) Use the substitution \(\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)\) and the answer to part (a) to prove that $$\frac { 1 - \sin x } { 1 + \sin x } \equiv ( \sec x - \tan x ) ^ { 2 } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$ \section*{Q uestion 1 continued}