1.3 Determine, without using a calculator, the value of the following in terms of \( p \), if \( \sin 43^{\circ}=p \) \( 1.3 .1 \quad \cos 133^{\circ} \) \( 1.3 .2 \tan \left(-43^{\circ}\right) \)

To determine \(\cos 133^{\circ}\), we can use the identity that relates cosines of angles. Notably, \(\cos(180^\circ - x) = -\cos x\). Here, \(133^\circ\) is in the second quadrant where cosine is negative. Therefore: \[ \cos(133^{\circ}) = \cos(180^{\circ} - 133^{\circ}) = -\cos(47^{\circ}) \] Since \(\sin 43^{\circ} = p\), we know \(\cos(47^{\circ}) = \sin(43^{\circ}) = p\) (because \(47^{\circ} + 43^{\circ} = 90^{\circ}\)). Thus, we have: \[ \cos(133^{\circ}) = -p \] For \( \tan(-43^{\circ}) \), we utilize the odd property of the tangent function: \(\tan(-x) = -\tan(x)\). Hence: \[ \tan(-43^{\circ}) = -\tan(43^{\circ}) \] Using the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), we find: \[ \tan(43^{\circ}) = \frac{\sin(43^{\circ})}{\cos(43^{\circ})} = \frac{p}{\cos(43^{\circ})} \] Given that \(\cos(43^{\circ}) = \sqrt{1 - p^2}\) (using \(\sin^2 + \cos^2 = 1\)), we can now express \(\tan(43^{\circ})\) as: \[ \tan(43^{\circ}) = \frac{p}{\sqrt{1 - p^2}} \] Finally, the value of \( \tan(-43^{\circ}) \) is: \[ \tan(-43^{\circ}) = -\frac{p}{\sqrt{1 - p^2}} \] In summary, the values are: 1.3.1 \(\cos 133^{\circ} = -p\) 1.3.2 \(\tan(-43^{\circ}) = -\frac{p}{\sqrt{1 - p^{2}}}\)