Given: \( \theta = 20^{\circ} \) and \( \alpha = 32^{\circ} \) Calculate the numerical value of: 3.1.1 \( \sin 3\alpha \) 3.1.2 \( \frac{\sec^2 \theta - 1}{\tan \alpha} \) - NSC Technical Mathematics - Question 3 - 2019 - Paper 2

For ( \sec^2 \theta ):
[ \sec^2 \theta = 1 + \tan^2 \theta ]
First, we calculate ( \tan \theta ) with ( \theta = 20^{\circ} ):
[ \tan 20^{\circ} \approx 0.36397 ]
Next, calculate ( \tan^2 \theta ):
[ \tan^2 20^{\circ} \approx 0.1325 ]
Thus,
[ \sec^2 20^{\circ} \approx 1 + 0.1325 = 1.1325 ]
Now we can substitute back into the expression:
[ \frac{1.1325 - 1}{\tan 32^{\circ}} ]
With ( \tan 32^{\circ} \approx 0.62488 ):
[ \frac{0.1325}{0.62488} \approx 0.21 ]