2 Simplify : \( \frac{\cos \left(\theta-90^{\circ}\right)}{\sin (90+\theta) \tan \left(720^{\circ}-\theta\right)} \) 3 If \( \cos 48^{\circ}=t \), determine in terms of \( t \) \( 5.3 .1 \cos 228^{\circ} \) \( 5.3 .2 \cos B+\sin \left(-42^{\circ}\right) \) \( 5.3 .3 \cos B, \cos 24^{\circ} \)

\( \textbf{Problem 2} \) We start with the expression \[ \frac{\cos\left(\theta-90^{\circ}\right)}{\sin(90^{\circ}+\theta) \tan\left(720^{\circ}-\theta\right)}. \] 1. Use the co‐function identity: \[ \cos\left(\theta-90^{\circ}\right) = \sin\theta. \] 2. Similarly, \[ \sin(90^{\circ}+\theta) = \cos\theta. \] 3. For the tangent term, note that \(720^{\circ}\) is a multiple of \(180^{\circ}\) (since \(720^{\circ} = 4 \times 180^{\circ}\)). Thus, \[ \tan\left(720^{\circ}-\theta\right) = \tan(-\theta). \] And since \[ \tan(-\theta) = -\tan\theta, \] we have \[ \tan\left(720^{\circ}-\theta\right) = -\tan\theta. \] Putting these results together, the expression becomes \[ \frac{\sin\theta}{\cos\theta \cdot (-\tan\theta)}. \] Since \[ \tan\theta = \frac{\sin\theta}{\cos\theta}, \] substitute to obtain \[ \frac{\sin\theta}{\cos\theta \cdot \left(-\frac{\sin\theta}{\cos\theta}\right)} = \frac{\sin\theta}{-\sin\theta} = -1. \] Thus, the simplified expression is \[ -1. \] --- \( \textbf{Problem 3} \) We are given that \[ \cos48^{\circ} = t. \] \( \textbf{5.3.1: } \cos228^{\circ} \) Notice that \[ 228^{\circ} = 180^{\circ} + 48^{\circ}. \] Using the cosine addition formula, \[ \cos(180^{\circ} + x) = -\cos x, \] we have \[ \cos228^{\circ} = -\cos48^{\circ} = -t. \] --- \( \textbf{5.3.2: } \cos B + \sin(-42^{\circ}) \) Recall that the sine function is odd: \[ \sin(-x) = -\sin x. \] Thus, \[ \sin(-42^{\circ}) = -\sin42^{\circ}. \] Moreover, using the co‐function identity, \[ \sin42^{\circ} = \cos(90^{\circ}-42^{\circ}) = \cos48^{\circ} = t. \] Therefore, \[ \cos B + \sin(-42^{\circ}) = \cos B - t. \] --- \( \textbf{5.3.3: } \cos B, \quad \cos24^{\circ} \) For \(\cos B\), there is no further simplification in terms of \(t\). For \(\cos24^{\circ}\), notice that \(24^{\circ}\) is half of \(48^{\circ}\). We use the half-angle formula for cosine: \[ \cos\frac{x}{2} = \sqrt{\frac{1+\cos x}{2}}, \] taking the positive square root since the angle \(\frac{x}{2} = 24^{\circ}\) is in the first quadrant. With \(x=48^{\circ}\) we get \[ \cos24^{\circ} = \sqrt{\frac{1+\cos48^{\circ}}{2}} = \sqrt{\frac{1+t}{2}}. \] --- \(\textbf{Summary of Answers}\) \[ \begin{aligned} \text{Problem 2:} \quad &\frac{\cos(\theta-90^{\circ})}{\sin(90^{\circ}+\theta) \tan(720^{\circ}-\theta)} = -1, \\ \\ \text{5.3.1:} \quad &\cos228^{\circ} = -t, \\ \\ \text{5.3.2:} \quad &\cos B + \sin(-42^{\circ}) = \cos B - t, \\ \\ \text{5.3.3:} \quad &\cos B \quad (\text{remains as is}), \quad \cos24^{\circ} = \sqrt{\frac{1+t}{2}}. \end{aligned} \]