3.1 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

To calculate the expression, we can use the Pythagorean identity:

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

Thus, we have:

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

Since we also know that:

$tan \beta = \frac{\sqrt{\frac{13}{9}}}{3} = \frac{\sqrt{13}}{3}$,

then,

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

Substituting into the expression gives:

$sec^2 \beta + tan^2 \beta = 1 + 2 \left(\frac{13}{9}\right) = 1 + \frac{26}{9} = \frac{9 + 26}{9} = \frac{35}{9}$

From the equation \cos \theta = \frac{1}{2}, we recognize that:

$\theta = 60^{\circ}$

Thus, the size of \theta is 60 degrees.

Given \tan \alpha = -1, we can deduce that in the second quadrant, the reference angle is:

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

Thus, the size of \alpha is 135 degrees.

Now, using the found values for \alpha and \theta:

\alpha = 135^{\circ}, \quad \theta = 60^{\circ}

Therefore:

$\cos(\alpha - \theta) = \cos(135^{\circ} - 60^{\circ}) = \cos(75^{\circ})$

Using the cosine of 75 degrees, which can be derived from angles:

$\cos(75^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4}$

Starting from the equation:

$2\tan x + 0.924 = 0$

Rearranging gives:

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

Since the tangent function is negative in the second and fourth quadrants:

For the second quadrant: $x = 180^{\circ} - \tan^{-1}(0.462)\approx 155.2^{\circ} \approx 155^{\circ}$

For the fourth quadrant: $x = 360^{\circ} - \tan^{-1}(0.462)\approx 335.2^{\circ} \approx 335^{\circ}$

Thus, the solutions for x are approximately: $x = 155^{\circ} ext{ and } x = 335^{\circ}$