6.1 Calculate the following without the use of a calculator: 6.2 Given that \( \tan A=-\frac{12}{5} \) and that \( A \) is not reflex, find the value of \( 35^{\circ}-\frac{\sin 90^{\circ}}{\cot 30^{\circ}} \) 6.3 \( \cos A \), without using a calculator.

To solve this, we first focus on \( 35^{\circ}-\frac{\sin 90^{\circ}}{\cot 30^{\circ}} \). We know that \( \sin 90^{\circ} = 1 \) and \( \cot 30^{\circ} = \frac{1}{\tan 30^{\circ}} = \frac{1}{\frac{\sqrt{3}}{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \). So we calculate: \[ \frac{\sin 90^{\circ}}{\cot 30^{\circ}} = \frac{1}{\sqrt{3}}. \] Then, \( 35^{\circ}-\frac{1}{\sqrt{3}} \) can be computed, but we need context for \( A \) first. Given \( \tan A = -\frac{12}{5} \), we can assume \( A \) lies in the second quadrant (since \( A \) is not reflex and tangent is negative there). From \( \tan A = \frac{\text{opposite}}{\text{adjacent}} \), we can establish a right triangle where the opposite side is 12 and the adjacent side is 5, allowing us to calculate \( \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \). Thus, for \( A \): \[ \sin A = \frac{12}{13}, \quad \cos A = -\frac{5}{13} \quad (\text{negative in the second quadrant}). \] Therefore, to find \( \cos A \) has led us to the result \( \cos A = -\frac{5}{13} \).