Given $Q = 42^{ ext{o}}$ and $P = 71^{ ext{o}}$\ Determine:\ 3.1.1 \text{ } cot(P - Q)\ 3.1.2 \text{ } \frac{cos Q}{sec P}\ \ Given $3sec \beta - 5 = 0$ and $\beta \in [90^{ ext{o}} ; 360^{ ext{o}}]$\ Determine $sin^{2} \beta - cos^{2} \beta$ with the aid of a diagram.\ \ Solve for $x$: $cos 2x - tan 29^{ ext{o}} = 0$ and $2x \in [0^{ ext{o}}; 360^{ ext{o}}]$ - NSC Technical Mathematics - Question 3 - 2022 - Paper 2

To find the ratio of $cos Q$ and $sec P$, we use the definitions of these trigonometric functions:

$sec P = \frac{1}{cos P}$ Thus, $\frac{cos Q}{sec P} = cos(42^{ ext{o}}) \times cos(71^{ ext{o}})$ Substituting the values: $\frac{cos(42^{ ext{o}})}{sec(71^{ ext{o}})} \approx 0.24$

From the equation $3sec \beta - 5 = 0$, we can isolate $sec \beta$:

$sec \beta = \frac{5}{3}$ The identity $sec \beta = \frac{1}{cos \beta}$ gives: $cos \beta = \frac{3}{5}$ Next, we calculate: $sin^{2} \beta = 1 - cos^{2} \beta = 1 - \left(\frac{3}{5}\right)^{2} = 1 - \frac{9}{25} = \frac{16}{25}$ Thus: $sin^{2} \beta - cos^{2} \beta = \frac{16}{25} - \left(\frac{3}{5}\right)^{2} = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$

To solve for $x$, start with:

$cos 2x = tan 29^{ ext{o}}$ Substituting the value: $cos 2x = 0.554309051$ Finding $2x$: $2x = cos^{-1}(0.554309051)$ Calculating: $2x \approx 34.34^{ ext{o}}$ Considering the periodicity of cosine functions, we find: $x \approx 17.17^{ ext{o}}$ or $x \approx 151.83^{ ext{o}}$