(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) \( \frac{\sin ^{2} 33^{\circ}+\sin ^{2} 123^{\circ}-\cos \left(-10^{\circ}\right) \cdot \sin 80^{\circ}}{\tan ^{2}\left(-150^{\circ}\right) \cdot \cos ^{2} 800^{\circ}} \)

Let's solve the problem step by step. ### Part (a) We start with the equation: \[ 5 \sin \theta - 4 = 0 \] **Step 1: Solve for \(\sin \theta\)** Rearranging gives: \[ 5 \sin \theta = 4 \implies \sin \theta = \frac{4}{5} \] **Step 2: Determine \(\theta\)** Given that \(90^{\circ} < \theta < 270^{\circ}\), \(\theta\) is in the second or third quadrant. In these quadrants, \(\sin \theta\) is positive in the second quadrant and negative in the third quadrant. Since \(\sin \theta = \frac{4}{5}\) is positive, we conclude: \[ \theta = 180^{\circ} - \alpha \quad \text{where } \sin \alpha = \frac{4}{5} \] Using the Pythagorean identity: \[ \cos^2 \alpha + \sin^2 \alpha = 1 \implies \cos^2 \alpha = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \implies \cos \alpha = \frac{3}{5} \] **Step 3: Find \(\cos \theta\) and \(\tan \theta\)** Since \(\theta = 180^{\circ} - \alpha\): \[ \cos \theta = -\cos \alpha = -\frac{3}{5} \] For \(\tan \theta\): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \] **Step 4: Calculate \(5 \cos \theta - 9 \tan^2 \theta\)** Now we compute: \[ 5 \cos \theta = 5 \left(-\frac{3}{5}\right) = -3 \] Next, we calculate \(\tan^2 \theta\): \[ \tan^2 \theta = \left(-\frac{4}{3}\right)^2 = \frac{16}{9} \] Now, calculate \(9 \tan^2 \theta\): \[ 9 \tan^2 \theta = 9 \cdot \frac{16}{9} = 16 \] Finally, we find: \[ 5 \cos \theta - 9 \tan^2 \theta = -3 - 16 = -19 \] Thus, the answer for part (a) is: \[ \boxed{-19} \] ### Part (b) **(1)** Simplify: \[ \sin(180^{\circ} - \theta) \cdot \cos(90^{\circ} - \theta) + \cos(-\theta) \cdot \sin(90^{\circ} + \theta) \] Using identities: \[ \sin(180^{\circ} - \theta) = \sin \theta, \quad \cos(90^{\circ} - \theta) = \sin \theta, \quad \cos(-\theta) = \cos \theta, \quad \sin(90^{\circ} + \theta) = \cos \theta \] Thus, we have: \[ \sin \theta \cdot \sin \theta + \cos \theta \cdot \cos \theta = \sin^2 \theta + \cos^2 \theta = 1 \] So, the answer for (1) is: \[ \boxed{1} \] **(2)** Simplify: \[ \frac{\tan(540^{\circ} + 1) \cdot \cos(360^{\circ} - x)}{\sin(180^{\circ} + x)} \] Using periodic properties: \[ \tan(540^{\circ} + 1) = \tan(180^{\circ} + 1) = \tan(1) \] \[ \cos(360^{\circ} - x) = \cos x \] \[ \sin(180^{\circ} + x) = -\sin x \] Thus, we have: \[ \frac{\tan(1) \cdot \cos x}{-\sin x} = -\tan(1) \cdot \cot x = -\frac{\tan(1)}{\tan x} \] So, the answer for (2) is: \[ \boxed{-\frac{\tan(1)}{\tan x}} \] **(3)** Simplify: \[ \frac{\cos(720^{\circ} + \theta) \cdot \sin(90^{\circ} - \theta)}{\tan(1800^{\circ} + \theta) \cdot \tan(90^{\circ} - \theta) - \sin(\theta - 180^{\circ}) \cdot \cos(90^{\circ} + \theta)} \] Using periodic properties: \[ \cos(720^{\circ} + \theta) = \cos \theta \] \[ \sin(90^{\circ} - \theta) = \cos \theta \] \[ \tan(1800^{\circ} + \theta) = \tan \theta \] \[ \tan(90^{\circ} - \theta) = \cot \theta \] \[ \sin(\theta - 180^{\circ}) = -\sin \theta \] \[ \cos(90^{\circ} + \theta) = -\sin \theta \] Thus, we have: \[ \frac{\cos \theta \cdot \cos \theta}{\tan \theta \cdot \cot \theta - (-\sin \theta)(-\sin \theta)} = \frac{\cos^2 \theta}{1 - \sin^2 \theta} = \frac{\cos^2 \theta}{\cos^2 \theta} = 1 \] So, the answer for (3) is: \[ \boxed{1} \] **(4)** Simplify: \[ \frac{\tan 45^{\circ} + \cos(180^{\circ} - \alpha) \cdot \sin(90^{\circ} + \alpha)}{\sin((\alpha - 360^{\circ}) \cdot \cos(90^{\circ} - \alpha) - \cos(180^{\circ} - \alpha) \cdot \cos(-\alpha))} \] Using identities: \[ \tan 45^{\circ} = 1, \quad \cos(180^{\circ} - \alpha) = -