3a) Proof the following trigonometric theorems.

Substitute this into the RHS.

$$\text{RHS} = \left(\frac{\sin \theta}{\cos \theta}\right) \cos \theta$$

$$\text{LHS} = \text{RHS}$$ $$\boxed{\sin \theta = \tan \theta \cos \theta}$$

ii) $\sec^2 \theta = 1 + \tan^2 \theta$

$$\left(\frac{\sin \theta}{\cos \theta}\right)^2 + 1 = \left(\frac{1}{\cos \theta}\right)^2$$ $$\tan^2 \theta + 1 = \sec^2 \theta$$

$$\boxed{\sec^2 \theta = 1 + \tan^2 \theta}$$

3b) State three (3) conditions each for using:

i) Sine Rule
The Sine Rule ($\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$) is used to solve triangles when you know:

• Two angles and any one side (AAS or ASA).
• Two sides and a non-included angle (SSA, which can sometimes lead to an ambiguous case with two possible triangles).
• To find the area of a triangle when two sides and the included angle are known (Area = $\frac{1}{2}ab \sin C$).

ii) Cosine Rule
The Cosine Rule ($a^2 = b^2 + c^2 - 2bc \cos A$) is used to solve triangles when you know:

• Two sides and the included angle (SAS).
• All three sides (SSS).
• To find the third side of a triangle when two sides and the included angle are known.

3c) Write down the values of the following using the triangle below.

Let the top vertex be A, the left base vertex be B, and the right base vertex be C. Let D be the point on BC where the altitude from A meets the base.
From the image, we have:
Altitude $AD = 22$
Base segment $BD = 11$
Base segment $DC = 14$
Angle at C is $x$. Angle at B is $y$.

First, we need to find the lengths of the hypotenuses in the two right-angled triangles formed by the altitude.

For $\triangle ADC$:
Using the Pythagorean theorem, $AC^2 = AD^2 + DC^2$.

$$AC^2 = 22^2 + 14^2$$ $$AC^2 = 484 + 196$$ $$AC^2 = 680$$ $$AC = \sqrt{680} = \sqrt{4 \times 170} = 2\sqrt{170}$$

For $\triangle ADB$:
Using the Pythagorean theorem, $AB^2 = AD^2 + BD^2$.

$$AB^2 = 22^2 + 11^2$$ $$AB^2 = 484 + 121$$ $$AB^2 = 605$$ $$AB = \sqrt{605} = \sqrt{121 \times 5} = 11\sqrt{5}$$

Now we can find the required trigonometric values:

i) $\cos x$
In right-angled $\triangle ADC$:

$$\cos x = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{DC}{AC}$$ $$\cos x = \frac{14}{2\sqrt{170}} = \frac{7}{\sqrt{170}}$$

To rationalize the denominator, multiply by $\frac{\sqrt{170}}{\sqrt{170}}$:

$$\cos x = \frac{7\sqrt{170}}{170}$$ $$\boxed{\cos x = \frac{7\sqrt{170}}{170}}$$

ii) $\sin y$
In right-angled $\triangle ADB$:

$$\sin y = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AD}{AB}$$ $$\sin y = \frac{22}{11\sqrt{5}} = \frac{2}{\sqrt{5}}$$

To rationalize the denominator, multiply by $\frac{\sqrt{5}}{\sqrt{5}}$:

$$\sin y = \frac{2\sqrt{5}}{5}$$ $$\boxed{\sin y = \frac{2\sqrt{5}}{5}}$$

iii) $\tan x$
In right-angled $\triangle ADC$:

$$\tan x = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AD}{DC}$$ $$\tan x = \frac{22}{14}$$ $$\tan x = \frac{11}{7}$$ $$\boxed{\tan x = \frac{11}{7}}$$