2. Calculate without using a calculator. a) \( \cos 30^{\circ} \times \sin 60^{\circ} \) b) \( \sec ^{2} 45^{\circ}-\tan ^{2} 45^{\circ} \) c) \( \sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ} \) d) \( 2 \cos ^{2} 45^{\circ}-1 \) e) \( \operatorname{cosec} 90^{\circ} \times \sin 90^{\circ} \) f) \( \cot ^{2} 45^{\circ}+\tan ^{2} 60^{\circ}-\sin ^{2} 0^{\circ}-\cos ^{2} 0^{\circ} \)

Alright, I need to calculate these trigonometric expressions without using a calculator. Let's tackle them one by one. **a) \( \cos 30^{\circ} \times \sin 60^{\circ} \)** First, I recall that \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) and \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \). Multiplying these together: \[ \cos 30^{\circ} \times \sin 60^{\circ} = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{3}{4} \] So, the result is \( \frac{3}{4} \). **b) \( \sec ^{2} 45^{\circ}-\tan ^{2} 45^{\circ} \)** I know that \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). For \( 45^{\circ} \), \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \). Calculating \( \sec 45^{\circ} \): \[ \sec 45^{\circ} = \frac{1}{\cos 45^{\circ}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \] Then, \( \sec^2 45^{\circ} = (\sqrt{2})^2 = 2 \). Next, \( \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 \). So, \( \tan^2 45^{\circ} = 1^2 = 1 \). Subtracting these: \[ \sec^2 45^{\circ} - \tan^2 45^{\circ} = 2 - 1 = 1 \] The result is \( 1 \). **c) \( \sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ} \)** Using the values I know: \[ \sin 60^{\circ} = \frac{\sqrt{3}}{2}, \quad \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \quad \cos 60^{\circ} = \frac{1}{2}, \quad \sin 30^{\circ} = \frac{1}{2} \] Plugging these in: \[ \sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ} = \left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \times \frac{1}{2} \right) = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \] So, the result is \( \frac{1}{2} \). **d) \( 2 \cos ^{2} 45^{\circ}-1 \)** First, \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \), so: \[ \cos^2 45^{\circ} = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{2}{4} = \frac{1}{2} \] Multiplying by 2: \[ 2 \cos^2 45^{\circ} = 2 \times \frac{1}{2} = 1 \] Subtracting 1: \[ 2 \cos^2 45^{\circ} - 1 = 1 - 1 = 0 \] The result is \( 0 \). **e) \( \operatorname{cosec} 90^{\circ} \times \sin 90^{\circ} \)** I recall that \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \). For \( 90^{\circ} \), \( \sin 90^{\circ} = 1 \), so: \[ \operatorname{cosec} 90^{\circ} = \frac{1}{1} = 1 \] Multiplying by \( \sin 90^{\circ} \): \[ \operatorname{cosec} 90^{\circ} \times \sin 90^{\circ} = 1 \times 1 = 1 \] The result is \( 1 \). **f) \( \cot ^{2} 45^{\circ}+\tan ^{2} 60^{\circ}-\sin ^{2} 0^{\circ}-\cos ^{2} 0^{\circ} \)** First, let's find each trigonometric value: - \( \cot 45^{\circ} = \frac{1}{\tan 45^{\circ}} = 1 \), so \( \cot^2 45^{\circ} = 1^2 = 1 \) - \( \tan 60^{\circ} = \sqrt{3} \), so \( \tan^2 60^{\circ} = (\sqrt{3})^2 = 3 \) - \( \sin 0^{\circ} = 0 \), so \( \sin^2 0^{\circ} = 0^2 = 0 \) - \( \cos 0^{\circ} = 1 \), so \( \cos^2 0^{\circ} = 1^2 = 1 \) Now, plugging these into the expression: \[ \cot^2 45^{\circ} + \tan^2 60^{\circ} - \sin^2 0^{\circ} - \cos^2 0^{\circ} = 1 + 3 - 0 - 1 = 3 \] The result is \( 3 \).