5.1 If \( \sqrt{3} \cdot \frac{1}{\cos \alpha}+2=0 \) and \( \tan \alpha<0 \), determine with the aid of a diagram the value of \( 10 \sin \alpha-3 \tan ^{2} \alpha \). 5.2 If \( \cos 12^{\circ}=m \), write down the following in terms of \( m \) : 5.2.1 \( \sin 168^{\circ} \)

1. **Problem 5.1** Given \[ \sqrt{3}\cdot\frac{1}{\cos \alpha}+2=0, \] we first isolate the term with \(\cos \alpha\): \[ \sqrt{3}\cdot\frac{1}{\cos \alpha}=-2 \quad \Longrightarrow \quad \frac{\sqrt{3}}{\cos \alpha}=-2. \] Multiply both sides by \(\cos \alpha\): \[ \sqrt{3}=-2\cos \alpha. \] Dividing by \(-2\) gives: \[ \cos \alpha=-\frac{\sqrt{3}}{2}. \] The condition \(\tan \alpha<0\) (and knowing \(\tan \alpha=\frac{\sin \alpha}{\cos \alpha}\)) implies that \(\sin \alpha\) and \(\cos \alpha\) have opposite signs. Since \(\cos \alpha<0\), it must be that \(\sin \alpha>0\). Thus, \(\alpha\) is in quadrant II. In quadrant II, using the Pythagorean identity: \[ \sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\left(-\frac{\sqrt{3}}{2}\right)^2}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}. \] Then, \[ \tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}. \] Now evaluate \[ 10\sin\alpha-3\tan^2\alpha. \] Substitute the values: \[ 10\left(\frac{1}{2}\right)-3\left(-\frac{\sqrt{3}}{3}\right)^2=5-3\left(\frac{3}{9}\right)=5-3\left(\frac{1}{3}\right)=5-1=4. \] **Diagram Aid:** - Draw a unit circle. - Locate quadrant II where \(\cos\alpha\) is negative and \(\sin\alpha\) is positive. - Mark the point on the circle corresponding to an angle where \(\cos \alpha = -\frac{\sqrt{3}}{2}\) and \(\sin \alpha = \frac{1}{2}\). **Conclusion:** The value of \(10\sin\alpha-3\tan^2\alpha\) is \(\boxed{4}\). 2. **Problem 5.2** Given \[ \cos 12^\circ = m, \] we need to express \(\sin 168^\circ\) in terms of \(m\). Recognize that \[ 168^\circ = 180^\circ - 12^\circ, \] so \[ \sin 168^\circ = \sin(180^\circ-12^\circ) = \sin 12^\circ. \] Using the Pythagorean identity: \[ \sin^2 12^\circ + \cos^2 12^\circ = 1, \] substitute \(\cos 12^\circ = m\): \[ \sin^2 12^\circ = 1 - m^2. \] Since \(12^\circ\) is in the first quadrant where sine is positive, we have: \[ \sin 12^\circ = \sqrt{1-m^2}. \] **Conclusion:** \(\sin 168^\circ = \sqrt{1-m^2}\).