4.1 Calculate $\cos B$ - NSC Technical Mathematics - Question 4 - 2023 - Paper 2

Using the identity, $\sin^2 P = 1 - \cos^2 P$.

We know that $\cot \beta = \frac{1}{\tan \beta}$, where $\tan \beta = \frac{\sin \beta}{\cos \beta}$.

Using $\sec^2 \beta - 1 = \tan^2 \beta$, we can rewrite as $\frac{\tan^2 \beta}{\sin^2 \beta} = \frac{1}{\cos^2 \beta}$.

We can leverage trigonometric identities: $\sin(\pi + B) = -\sin B$, $\cos(2\pi - B) = \cos B$ and $\cos(180 - B) = -\cos B$. Therefore, the left-hand side simplifies to $2(-\sin B \cos B - \cos B) = 2 \sec(180+B)$.

By rearranging both sides, we can express $\tan \theta = \frac{\sin \theta}{\cos \theta}$. Thus, validating both sides leads to the equality.