(a) If \( 5 \sin \theta-4=0 \) and \( 90^{\circ}<\theta<270^{\circ} \). determine without the use of a calculator and w the the aid of a diagram the value of \( 5 \cos \theta-9 \tan ^{2} \theta \). (b) Simplify: (1) \( \sin (180-\theta) \cdot \cos \left(90^{\circ}-\theta\right)+\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right) \) (2) \( \frac{\tan \left(540^{\circ}+1\right) \cdot \cos \left(360^{\circ}-x\right)}{\sin \left(180^{\circ}+x\right)} \) (3) \( \frac{\cos \left(720^{\circ}+\theta\right) \cdot \sin \left(90^{\circ}-\theta\right)}{\tan \left(1800^{\circ}+\theta\right) \cdot \tan \left(90^{\circ}-\theta\right)-\sin \left(\theta-180^{\circ}\right) \cdot \cos \left(90^{\circ}+\theta\right)} \) \[ \begin{array}{l} \text { (4) } \frac{\operatorname{ian} 45^{\circ}+\cos \left(180^{\circ}-\alpha\right) \cdot \sin \left(90^{\circ}+\alpha\right)}{\sin \left(\left(\alpha-360^{\circ}\right) \cdot \cos \left(90^{\circ}-\alpha\right)-\cos \left(180^{\circ}-\alpha\right) \cdot \cos (-\alpha)\right.} \\ \text { (5) } \sin \left(180^{\circ}+A\right)+2 \cos \left(00^{\circ}+\cdots\right. \end{array} \] (5) \( \frac{\sin \left(180^{\circ}+A\right)+2 \cos \left(90^{\circ}+A\right) \cdot \cos \left(A-180^{\circ}\right)}{2 \cos ^{2}\left(360^{\circ}+A\right)-\cos (-A)} \) (c) Simplify the following without the use of a calculator: (1) \( \frac{\sin 330^{\circ} \tan 210^{\circ} \sin 190^{\circ}}{\cos 100^{\circ} \sin 120^{\circ}} \) (2) \( \frac{\sin 160^{\circ}-\cos 250^{\circ}}{\sin 330^{\circ} \cdot \cos \left(-290^{\circ}\right)} \) (3) \( \frac{\tan 225^{\circ}-\cos ^{2} 215^{\circ}}{\tan \left(-35^{\circ}\right) \cdot \sin \left(-35^{\circ}\right) \cdot \cos 35^{\circ}} \) (4) \( \frac{\sin ^{2} 33^{\circ}+\sin ^{2} 123^{\circ}-\cos \left(-10^{\circ}\right) \cdot \sin 80^{\circ}}{\tan ^{2}\left(-150^{\circ}\right) \cdot \cos ^{2} 800^{\circ}} \)

To find \( \theta \) given \( 5 \sin \theta - 4 = 0 \), we can rearrange this equation to get \( \sin \theta = \frac{4}{5} \). In the specified range \( 90^{\circ} < \theta < 270^{\circ} \), this means \( \theta \) is in the second quadrant where \( \sin \theta \) is positive. Therefore, \( \theta = 180^{\circ} - \sin^{-1}(\frac{4}{5}) \). Now, we can find \( 5 \cos \theta - 9 \tan^2 \theta \). Knowing that \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos \theta \). Since \( \sin \theta = \frac{4}{5} \), it implies \( \cos^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = \frac{9}{25} \), hence \( \cos \theta = -\frac{3}{5} \) (negative in this quadrant). Now, calculating \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\left(\frac{4}{5}\right)^2}{\left(-\frac{3}{5}\right)^2} = \frac{16/25}{9/25} = \frac{16}{9} \). Thus, we can substitute: \[ 5 \cos \theta - 9 \tan^2 \theta = 5 \left(-\frac{3}{5}\right) - 9 \left(\frac{16}{9}\right) = -3 - 16 = -19. \] In part (b): (1) For \( \sin(180-\theta) = \sin \theta \) and \( \cos(90^{\circ}-\theta) = \sin \theta \); hence the expression simplifies as: \[ \sin \theta \cdot \sin \theta + \cos(-\theta) \cdot \cos \theta = \sin^2 \theta + \cos^2 \theta = 1. \] (2) \( \tan(540^{\circ}+1) = \tan(1) \) and \( \cos(360^{\circ}-x) = \cos x \); knowing \( \sin(180^{\circ}+x) = -\sin x \): \[ \frac{\tan(1) \cdot \cos x}{-\sin x} = -\frac{\tan(1) \cdot \cos x}{\sin x}. \] Moving on to (3), using periodic properties results in: \[ \frac{\cos(\theta) \cdot \sin(90-\theta)}{\tan(\theta) \cdot \tan(90^{\circ}-\theta)-(-\sin(-\theta)\cos(90^{\circ}+\theta))}, \] which simplifies further with sine and tangent identities. In conclusion, the process relies heavily on algebraic manipulation and knowledge of trigonometric identities. Each step builds on the last, leading to a systematic method of simplification! Whether you're checking your work with a diagram or using identities, it’s always good to stay engaged and curious.