In the diagram below, P(-7; 4) is a point in the Cartesian plane - NSC Mathematics - Question 5 - 2022 - Paper 2

To find tan θ, we can use the definition of tangent:

$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

Here, opposite = 4 and adjacent = -7. Therefore:

$\tan(\theta) = \frac{4}{-7} = -\frac{4}{7}$.

Using the cosine reduction formula:

$\cos(\theta - 180°) = -\cos(\theta)$

From previous calculations, we have:

$\cos(\theta) = \frac{-7}{\sqrt{65}}$

So,

$\cos(\theta - 180°) = -\left( -\frac{7}{\sqrt{65}} \right) = \frac{7}{\sqrt{65}}$.

We can rewrite the equation:

$\sin x (\cos x + 1) = 3 \cos x (\cos^2 x + 1)$.

Setting both sides to zero, we factor:

1. For \sin x = 0, ( x = k \cdot 180°, k ∈ Z )

2. For \cos x = 0, ( x = 90° + k \cdot 180°, k ∈ Z )

Combine solutions from both cases.

To prove this identity, start with the left-hand side:

$LHS = \frac{\sin 3x}{1 - \cos 3x} \cdot \frac{\sin 3x}{\sin 3x} = \frac{\sin^2 3x}{\sin 3x (1 - \cos 3x)}$.

Now, rewrite the expression using the Pythagorean identity, which leads to:

$= \frac{1 - \cos^2 3x}{\sin 3x (1 - \cos 3x)} = \frac{(1 - \cos 3x)(1 + \cos 3x)}{\sin 3x (1 - \cos 3x)} = \frac{1 + \cos 3x}{\sin 3x} = RHS$.

The identity will be undefined when:

1. $\sin 3x = 0 \Rightarrow 3x = k \cdot 180° \Rightarrow x = k \cdot 60°$

   • Valid solutions: $x = 0°$, $x = 60°$

2. $1 - \cos 3x = 0 \Rightarrow \cos 3x = 1 \Rightarrow 3x = 0° + k \cdot 360° \Rightarrow x = 0° + k \cdot 120°$

   • Valid solution: $x = 0°$.

Thus, $x = 0°$ is the only solution in the interval [0°, 60°] which makes the identity undefined.