Question 1 1. \( \sqrt{3} \operatorname{cosec} \theta+2=0 \) and \( 90<\theta<270 \) use sketch to determine the value of the following wilhout using a (denlocor i.1.) \( \cos \theta \) \( 1.1 .2 \frac{\tan \theta . \sec \theta}{\cos \theta} \) 1.2. Simplity withoce a colculntor \[ \frac{\sin 90-\cos 60}{\tan 45} \] 1.3 if \( 5 \sin x=4 \) and \( x \in[0,90] \) use a sketch to colculate the value of the following \( \sin ^{2} x+\cos ^{2} x \)

\( \textbf{Part 1.1:} \) We are given \[ \sqrt{3}\,\csc \theta + 2 = 0. \] Since \(\csc \theta = \frac{1}{\sin \theta}\), we write \[ \sqrt{3}\,\frac{1}{\sin \theta} + 2 = 0. \] Subtracting 2 from both sides gives \[ \frac{\sqrt{3}}{\sin \theta} = -2. \] Multiplying both sides by \(\sin \theta\) and then dividing by \(-2\) yields \[ \sin \theta = -\frac{\sqrt{3}}{2}. \] The reference angle corresponding to \(\sin \theta = \frac{\sqrt{3}}{2}\) is \(60^\circ\). Given that \(90^\circ < \theta < 270^\circ\), the sine is negative in the third quadrant. Thus, \[ \theta = 180^\circ + 60^\circ = 240^\circ. \] \(\textbf{i.1) Compute } \cos \theta:\) For \(\theta = 240^\circ\), \[ \cos 240^\circ = -\cos 60^\circ = -\frac{1}{2}. \] \(\textbf{i.1.2) Compute } \frac{\tan \theta \, \sec \theta}{\cos \theta}:\) We have \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta}. \] Thus, \[ \tan \theta \, \sec \theta = \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos \theta} = \frac{\sin \theta}{\cos^2 \theta}. \] Dividing by \(\cos \theta\) gives \[ \frac{\tan \theta \, \sec \theta}{\cos \theta} = \frac{\sin \theta}{\cos^3 \theta}. \] Substitute the known values: \[ \sin 240^\circ = -\frac{\sqrt{3}}{2}, \quad \cos 240^\circ = -\frac{1}{2}. \] Then, \[ \cos^3 240^\circ = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}. \] So, \[ \frac{\tan \theta \, \sec \theta}{\cos \theta} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{8}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{8}} = \frac{\sqrt{3}}{2} \times 8 = 4\sqrt{3}. \] \( \textbf{Part 1.2:} \) Simplify \[ \frac{\sin 90 - \cos 60}{\tan 45}. \] Evaluate each trigonometric function: \[ \sin 90^\circ = 1, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 45^\circ = 1. \] Thus, the numerator is \[ 1 - \frac{1}{2} = \frac{1}{2}, \] and the expression becomes \[ \frac{\frac{1}{2}}{1} = \frac{1}{2}. \] \( \textbf{Part 1.3:} \) We are given \[ 5\sin x = 4 \quad \text{with} \quad x \in [0^\circ, 90^\circ]. \] The expression to evaluate is \[ \sin^2 x + \cos^2 x. \] By the Pythagorean identity, for any angle \(x\), \[ \sin^2 x + \cos^2 x = 1. \] \(\boxed{ \begin{aligned} \text{1.1.1) } \cos \theta &= -\frac{1}{2}, \\ \text{1.1.2) } \frac{\tan \theta \, \sec \theta}{\cos \theta} &= 4\sqrt{3}, \\ \text{1.2) } \frac{\sin 90^\circ - \cos 60^\circ}{\tan 45^\circ} &= \frac{1}{2}, \\ \text{1.3) } \sin^2 x+\cos^2 x &= 1. \end{aligned} } \)