In the diagram below, P (3 ; m) is a point in a Cartesian plane with OP = \sqrt{13} \beta is an acute angle - NSC Technical Mathematics - Question 3 - 2021 - Paper 2

Using the identity for secant and tangent:

$sec^2 \beta = 1 + tan^2 \beta$

This means:

$sec^2 \beta + tan^2 \beta = (1 + tan^2 \beta) + tan^2 \beta = 1 + 2tan^2 \beta$

Next we substitute the value of $tan \beta$. Letting $tan \beta = \frac{\sqrt{13}}{3}$:

$tan^2 \beta = \left(\frac{\sqrt{13}}{3}\right)^2 = \frac{13}{9}$

Therefore:

$sec^2 \beta + tan^2 \beta = 1 + 2 \cdot \frac{13}{9} = 1 + \frac{26}{9} = \frac{9}{9} + \frac{26}{9} = \frac{35}{9}$

Given that \cos \theta = \frac{1}{2}$, we can use the inverse cosine function:

$\theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$

Here, we have \tan \alpha = -1. The reference angle where tangent is -1 is \alpha = 45^\circ$. Since \alpha must be in the second quadrant:

$\alpha = 180^\circ - 45^\circ = 135^\circ$

We found earlier that \alpha = 135^\circ and \theta = 60^\circ. Therefore:

$\alpha - \theta = 135^\circ - 60^\circ = 75^\circ$

Now calculating:

$\cos(75^\circ)$

Using the cosine angle subtraction identity:

$\cos(\alpha - \theta) = \cos \alpha \cos \theta + \sin \alpha \sin \theta$

Substituting the values where:

$\cos(135^\circ) = -\frac{\sqrt{2}}{2}, \sin(135^\circ) = \frac{\sqrt{2}}{2}, \cos(60^\circ) = \frac{1}{2}, \sin(60^\circ) = \frac{\sqrt{3}}{2}$

Thus, we obtain:

$\cos(75^\circ) = \left(-\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{-\sqrt{2} + \sqrt{6}}{4}$

Starting with the equation:

$2 \tan x + 0.924 = 0$

Rearranging gives:

$\tan x = -\frac{0.924}{2} = -0.462$

The general solution for tangent is given as:

$x = \tan^{-1}(-0.462) + n\cdot180^\circ$

Identifying the reference angle:

$x \approx 24^\circ$ (in quadrant 4)

Thus:

$x = 180^\circ - 24^\circ = 156^\circ$

And, for the angle in the fourth quadrant:

$x \approx 360^\circ - | an^{-1}(-0.462)| = 360^\circ - 24^\circ = 335^\circ$

Final answers are:

$x = 156^\circ, \; 335^\circ$