Prove identity then solve equation

7

1. Prove the identity \(\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } \equiv 1 - \tan ^ { 2 } \theta\).
2. Hence solve the equation \(\frac { 1 - 2 \sin ^ { 2 } \theta } { 1 - \sin ^ { 2 } \theta } = 2 \tan ^ { 4 } \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

10

1. Prove the identity \(\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } \equiv \frac { 4 \tan x } { \cos x }\).
2. Hence solve the equation \(\frac { 1 + \sin x } { 1 - \sin x } - \frac { 1 - \sin x } { 1 + \sin x } = 8 \tan x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

4

1. Show that the equation $$\frac { \tan x + \sin x } { \tan x - \sin x } = k$$ where \(k\) is a constant, may be expressed as $$\frac { 1 + \cos x } { 1 - \cos x } = k$$
2. Hence express \(\cos x\) in terms of \(k\).
3. Hence solve the equation \(\frac { \tan x + \sin x } { \tan x - \sin x } = 4\) for \(- \pi < x < \pi\).

4

1. Prove the identity \(\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } \equiv - \tan ^ { 2 } \theta \left( 1 + \sin ^ { 2 } \theta \right)\).
2. Hence solve the equation $$\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } = \tan ^ { 2 } \theta \left( 1 - \sin ^ { 2 } \theta \right)$$ for \(0 < \theta < 2 \pi\).

7

1. Show that \(\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } \equiv \frac { 4 } { 5 \cos ^ { 2 } \theta - 4 }\).
2. Hence solve the equation \(\frac { \sin \theta + 2 \cos \theta } { \cos \theta - 2 \sin \theta } - \frac { \sin \theta - 2 \cos \theta } { \cos \theta + 2 \sin \theta } = 5\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

7

1. Show that \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } \equiv 2 \tan ^ { 2 } \theta\).
2. Hence solve the equation \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } = 8\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

8

1. Prove the identity \(\left( \frac { 1 } { \sin \theta } - \frac { 1 } { \tan \theta } \right) ^ { 2 } \equiv \frac { 1 - \cos \theta } { 1 + \cos \theta }\).
2. Hence solve the equation \(\left( \frac { 1 } { \sin \theta } - \frac { 1 } { \tan \theta } \right) ^ { 2 } = \frac { 2 } { 5 }\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

5

1. Show that \(\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } \equiv \frac { 1 } { \sin ^ { 2 } \theta - \cos ^ { 2 } \theta }\).
2. Hence solve the equation \(\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } = 3\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

9

1. Prove the identity \(\frac { \sin \theta } { 1 - \cos \theta } - \frac { 1 } { \sin \theta } \equiv \frac { 1 } { \tan \theta }\).
2. Hence solve the equation \(\frac { \sin \theta } { 1 - \cos \theta } - \frac { 1 } { \sin \theta } = 4 \tan \theta\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

5

1. Prove the identity \(\frac { 1 } { \cos \theta } - \frac { \cos \theta } { 1 + \sin \theta } \equiv \tan \theta\).
2. Solve the equation \(\frac { 1 } { \cos \theta } - \frac { \cos \theta } { 1 + \sin \theta } + 2 = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

4

1. Prove the identity \(\frac { \tan x + 1 } { \sin x \tan x + \cos x } \equiv \sin x + \cos x\).
2. Hence solve the equation \(\frac { \tan x + 1 } { \sin x \tan x + \cos x } = 3 \sin x - 2 \cos x\) for \(0 \leqslant x \leqslant 2 \pi\).

7

1. Prove the identity \(\frac { 1 + \cos \theta } { 1 - \cos \theta } - \frac { 1 - \cos \theta } { 1 + \cos \theta } \equiv \frac { 4 } { \sin \theta \tan \theta }\).
2. Hence solve, for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), the equation $$\sin \theta \left( \frac { 1 + \cos \theta } { 1 - \cos \theta } - \frac { 1 - \cos \theta } { 1 + \cos \theta } \right) = 3 .$$

3

1. Prove the identity \(\left( \frac { 1 } { \cos \theta } - \tan \theta \right) ^ { 2 } \equiv \frac { 1 - \sin \theta } { 1 + \sin \theta }\).
2. Hence solve the equation \(\left( \frac { 1 } { \cos \theta } - \tan \theta \right) ^ { 2 } = \frac { 1 } { 2 }\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\). \includegraphics[max width=\textwidth, alt={}, center]{8a3f8707-67a4-4069-aba5-7e9496cb1748-06_572_460_258_845} The diagram shows a circle with radius \(r \mathrm {~cm}\) and centre \(O\). Points \(A\) and \(B\) lie on the circle and \(A B C D\) is a rectangle. Angle \(A O B = 2 \theta\) radians and \(A D = r \mathrm {~cm}\).

4

1. Prove the identity \(( \sin \theta + \cos \theta ) ( 1 - \sin \theta \cos \theta ) \equiv \sin ^ { 3 } \theta + \cos ^ { 3 } \theta\).
2. Hence solve the equation \(( \sin \theta + \cos \theta ) ( 1 - \sin \theta \cos \theta ) = 3 \cos ^ { 3 } \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

6

1. Prove the identity \(\left( \frac { 1 } { \cos x } - \tan x \right) ^ { 2 } \equiv \frac { 1 - \sin x } { 1 + \sin x }\).
2. Hence solve the equation \(\left( \frac { 1 } { \cos 2 x } - \tan 2 x \right) ^ { 2 } = \frac { 1 } { 3 }\) for \(0 \leqslant x \leqslant \pi\).

5

1. Prove the identity \(( \sin x + \cos x ) ( 1 - \sin x \cos x ) \equiv \sin ^ { 3 } x + \cos ^ { 3 } x\).
2. Solve the equation \(( \sin x + \cos x ) ( 1 - \sin x \cos x ) = 9 \sin ^ { 3 } x\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

4

1. Prove the identity \(\frac { \sin x \tan x } { 1 - \cos x } \equiv 1 + \frac { 1 } { \cos x }\).
2. Hence solve the equation \(\frac { \sin x \tan x } { 1 - \cos x } + 2 = 0\), for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

4

1. Prove the identity \(\left( \frac { 1 } { \sin x } - \frac { 1 } { \tan x } \right) ^ { 2 } \equiv \frac { 1 - \cos x } { 1 + \cos x }\).
2. Hence solve the equation \(\left( \frac { 1 } { \sin x } - \frac { 1 } { \tan x } \right) ^ { 2 } = \frac { 2 } { 5 }\) for \(0 \leqslant x \leqslant 2 \pi\).

7

1. Show that \(\frac { \tan \theta + 1 } { 1 + \cos \theta } + \frac { \tan \theta - 1 } { 1 - \cos \theta } \equiv \frac { 2 ( \tan \theta - \cos \theta ) } { \sin ^ { 2 } \theta }\).
2. Hence, showing all necessary working, solve the equation $$\frac { \tan \theta + 1 } { 1 + \cos \theta } + \frac { \tan \theta - 1 } { 1 - \cos \theta } = 0$$ for \(0 ^ { \circ } < \theta < 90 ^ { \circ }\).

9. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Given that \(\cos 2 \theta - \sin 3 \theta \neq 0\)

1. prove that $$\frac { \cos ^ { 2 } \theta } { \cos 2 \theta - \sin 3 \theta } \equiv \frac { 1 + \sin \theta } { 1 - 2 \sin \theta - 4 \sin ^ { 2 } \theta }$$
2. Hence solve, for \(0 < \theta \leqslant 360 ^ { \circ }\) $$\frac { \cos ^ { 2 } \theta } { \cos 2 \theta - \sin 3 \theta } = 2 \operatorname { cosec } \theta$$ Give your answers to one decimal place. \includegraphics[max width=\textwidth, alt={}, center]{83e12fa4-1abb-4bea-bff4-8d36757bd9c3-28_2257_52_309_1983}

6. (a) Use the double angle formulae and the identity $$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B$$ to obtain an expression for \(\cos 3 x\) in terms of powers of \(\cos x\) only.
(b) (i) Prove that $$\frac { \cos x } { 1 + \sin x } + \frac { 1 + \sin x } { \cos x } \equiv 2 \sec x , \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 }$$ (ii) Hence find, for \(0 < x < 2 \pi\), all the solutions of $$\frac { \cos x } { 1 + \sin x } + \frac { 1 + \sin x } { \cos x } = 4$$

1. (a) Prove that

$$\frac { \sin \theta } { \cos \theta } + \frac { \cos \theta } { \sin \theta } = 2 \operatorname { cosec } 2 \theta , \quad \theta \neq 90 n ^ { \circ }$$ (b) On the axes on page 20, sketch the graph of \(y = 2 \operatorname { cosec } 2 \theta\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
(c) Solve, for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), the equation $$\frac { \sin \theta } { \cos \theta } + \frac { \cos \theta } { \sin \theta } = 3 ,$$ giving your answers to 1 decimal place. \includegraphics[max width=\textwidth, alt={}, center]{f3c3c777-7808-4d82-a1f4-2dee6674be1e-11_899_1253_315_347}

1. (a) Show that

$$\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta$$ (b) Hence find, for \(- 180 ^ { \circ } \leqslant \theta < 180 ^ { \circ }\), all the solutions of $$\frac { 2 \sin 2 \theta } { 1 + \cos 2 \theta } = 1$$ Give your answers to 1 decimal place.

6. (a) Prove that $$\frac { 1 } { \sin 2 \theta } - \frac { \cos 2 \theta } { \sin 2 \theta } = \tan \theta , \quad \theta \neq 90 n ^ { \circ } , n \in \mathbb { Z }$$ (b) Hence, or otherwise,

1. show that \(\tan 15 ^ { \circ } = 2 - \sqrt { 3 }\),
2. solve, for \(0 < x < 360 ^ { \circ }\), $$\operatorname { cosec } 4 x - \cot 4 x = 1$$

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 $$\frac { \cos x } { 1 - \sin x } + \frac { 1 - \sin x } { \cos x } = 8 \sin x$$