(f) Given that \( \cos \theta=t \), with \( -90^{\circ}<\theta<0^{\circ} \). (1) What is the range of possible values of \( t \) ? (2) Express the following in terms of \( t \) : \( \begin{array}{lll}\text { (i) } \tan \left(180^{\circ}+\theta\right) & \text { (ii) } \sin 2 \theta & \text { (iii) } \cos \left(60^{\circ}+\theta\right)\end{array} \)

**(1) Range of \( t \):** Since \[ \cos \theta = t \quad \text{and} \quad -90^\circ < \theta < 0^\circ, \] we note that when \(-90^\circ < \theta < 0^\circ\) the angle \(\theta\) lies in the fourth quadrant, where the cosine function is positive. As \(\theta\) approaches \(0^\circ\), \(\cos \theta\) approaches \(1\), and as \(\theta\) approaches \(-90^\circ\), \(\cos \theta\) approaches \(0\) (but not including \(0\) or \(1\) since the endpoints are not included). Thus, the range of \( t \) is \[ 0 < t < 1. \] **(2) Expressing in terms of \( t \):** *First, note the identity:* \[ \sin^2\theta + \cos^2\theta = 1 \quad \text{or} \quad \sin^2\theta = 1 - t^2. \] Since \(\theta\) is in the fourth quadrant, \(\sin \theta\) is negative. Therefore, \[ \sin \theta = -\sqrt{1-t^2}. \] **(i) \(\tan \left(180^\circ+\theta\right)\):** Using the periodic property of the tangent function, \[ \tan \left(180^\circ+\theta\right)=\tan \theta. \] Thus, \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\sqrt{1-t^2}}{t}. \] So, \[ \tan \left(180^\circ+\theta\right) = -\frac{\sqrt{1-t^2}}{t}. \] **(ii) \(\sin 2\theta\):** Using the double-angle formula, \[ \sin 2\theta = 2\sin \theta \cos \theta. \] Substitute \(\sin \theta = -\sqrt{1-t^2}\) and \(\cos \theta = t\) to get: \[ \sin 2\theta = 2t(-\sqrt{1-t^2}) = -2t\sqrt{1-t^2}. \] **(iii) \(\cos \left(60^\circ+\theta\right)\):** Using the cosine addition formula, \[ \cos \left(60^\circ+\theta\right)=\cos 60^\circ \cos \theta - \sin 60^\circ \sin \theta. \] Recall that \(\cos 60^\circ=\frac{1}{2}\) and \(\sin 60^\circ=\frac{\sqrt{3}}{2}\). Substituting, we have: \[ \cos \left(60^\circ+\theta\right)=\frac{1}{2}t - \frac{\sqrt{3}}{2}(-\sqrt{1-t^2}) =\frac{1}{2}t+\frac{\sqrt{3}}{2}\sqrt{1-t^2}. \] **Final Answers:** 1. The range of \( t \) is \(\quad 0 < t < 1 \). 2. In terms of \( t \): - (i) \(\displaystyle \tan \left(180^\circ+\theta\right) = -\frac{\sqrt{1-t^2}}{t}\). - (ii) \(\displaystyle \sin 2\theta = -2t\sqrt{1-t^2}\). - (iii) \(\displaystyle \cos \left(60^\circ+\theta\right) = \frac{1}{2}\left(t+\sqrt{3}\sqrt{1-t^2}\right)\).