Use the special angles to evaluate: a) \( \operatorname{cosec} 60^{\circ} \cdot \cot 30^{\circ}+\cos 45^{\circ} \cdot \operatorname{cosec} 45^{\circ} \) b) \( \sec ^{2} 45^{\circ}-\tan ^{2} 45^{\circ} \) c) \( 1+\cot ^{2} 60^{\circ} \) d) \( \operatorname{cosec}^{2} 60^{\circ}+\cot ^{2} 45^{\circ}+\sec ^{2} 30^{\circ} \) e) \( \sin 90^{\circ}-\cos 60^{\circ} \cdot \sec 60^{\circ}+\operatorname{cosec} 45^{\circ} \cdot \cos 45^{\circ} \) f) \( \operatorname{cosec} 60^{\circ} \cdot \tan 45^{\circ} \cdot \sin 30^{\circ} \)

a) We start with \[ \csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}, \quad \cot 30^\circ = \frac{\cos 30^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}. \] Then, \[ \csc 60^\circ \cdot \cot 30^\circ = \frac{2}{\sqrt{3}} \cdot \sqrt{3} = 2. \] Next, we have \[ \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}. \] Thus, \[ \cos 45^\circ \cdot \csc 45^\circ = \frac{\sqrt{2}}{2} \cdot \sqrt{2} = \frac{2}{2} = 1. \] Adding the two parts, \[ \csc 60^\circ \cdot \cot 30^\circ + \cos 45^\circ \cdot \csc 45^\circ = 2 + 1 = 3. \] --- b) We know that \[ \tan 45^\circ = 1, \quad \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}. \] Thus, \[ \sec^2 45^\circ = (\sqrt{2})^2 = 2, \quad \tan^2 45^\circ = 1. \] Therefore, \[ \sec^2 45^\circ - \tan^2 45^\circ = 2 - 1 = 1. \] --- c) We have \[ \cot 60^\circ = \frac{\cos 60^\circ}{\sin 60^\circ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}, \] so \[ \cot^2 60^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}. \] Thus, \[ 1 + \cot^2 60^\circ = 1 + \frac{1}{3} = \frac{4}{3}. \] --- d) Compute each term separately: 1. For \( \csc^2 60^\circ \): \[ \csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{2}{\sqrt{3}}, \quad \csc^2 60^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}. \] 2. For \( \cot^2 45^\circ \): \[ \cot 45^\circ = 1 \quad \Rightarrow \quad \cot^2 45^\circ = 1. \] 3. For \( \sec^2 30^\circ \): \[ \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{2}{\sqrt{3}}, \quad \sec^2 30^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}. \] Now, adding all three terms: \[ \csc^2 60^\circ + \cot^2 45^\circ + \sec^2 30^\circ = \frac{4}{3} + 1 + \frac{4}{3} = 1 + \frac{8}{3} = \frac{11}{3}. \] --- e) Evaluate each term: 1. \(\sin 90^\circ = 1.\) 2. For \(\cos 60^\circ \cdot \sec 60^\circ\): \[ \cos 60^\circ = \frac{1}{2}, \quad \sec 60^\circ = \frac{1}{\cos 60^\circ} = 2, \quad \cos 60^\circ \cdot \sec 60^\circ = \frac{1}{2} \cdot 2 = 1. \] 3. For \(\csc 45^\circ \cdot \cos 45^\circ\): \[ \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \csc 45^\circ \cdot \cos 45^\circ = \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{2} = 1. \] Thus, \[ \sin 90^\circ - \cos 60^\circ \cdot \sec 60^\circ + \csc 45^\circ \cdot \cos 45^\circ = 1 - 1 + 1 = 1. \] --- f) Compute each factor: 1. \(\csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{2}{\sqrt{3}}.\) 2. \(\tan 45^\circ = 1.\) 3. \(\sin 30^\circ = \frac{1}{2}.\) Multiplying the three factors: \[ \csc 60^\circ \cdot \tan 45^\circ \cdot \sin 30^\circ = \frac{2}{\sqrt{3}} \cdot 1 \cdot \frac{1}{2} = \frac{1}{\sqrt{3}}. \]