b. \( 3 \tan ^{2} 45^{\circ}-\sin ^{2} 60^{\circ}-\frac{1}{2} \cot ^{2} 30^{\circ}+\frac{1}{8} \sec ^{2} 45^{\circ} \)

To evaluate the expression \[ 3 \tan^{2} 45^{\circ} - \sin^{2} 60^{\circ} - \frac{1}{2} \cot^{2} 30^{\circ} + \frac{1}{8} \sec^{2} 45^{\circ}, \] we will calculate each trigonometric function step by step. 1. **Calculate \( \tan 45^{\circ} \)**: \[ \tan 45^{\circ} = 1 \implies \tan^{2} 45^{\circ} = 1^{2} = 1. \] Therefore, \[ 3 \tan^{2} 45^{\circ} = 3 \cdot 1 = 3. \] 2. **Calculate \( \sin 60^{\circ} \)**: \[ \sin 60^{\circ} = \frac{\sqrt{3}}{2} \implies \sin^{2} 60^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4}. \] 3. **Calculate \( \cot 30^{\circ} \)**: \[ \cot 30^{\circ} = \frac{1}{\tan 30^{\circ}} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \implies \cot^{2} 30^{\circ} = (\sqrt{3})^{2} = 3. \] Therefore, \[ -\frac{1}{2} \cot^{2} 30^{\circ} = -\frac{1}{2} \cdot 3 = -\frac{3}{2}. \] 4. **Calculate \( \sec 45^{\circ} \)**: \[ \sec 45^{\circ} = \frac{1}{\cos 45^{\circ}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \implies \sec^{2} 45^{\circ} = (\sqrt{2})^{2} = 2. \] Therefore, \[ \frac{1}{8} \sec^{2} 45^{\circ} = \frac{1}{8} \cdot 2 = \frac{2}{8} = \frac{1}{4}. \] Now, we can substitute all these values back into the original expression: \[ 3 - \frac{3}{4} - \frac{3}{2} + \frac{1}{4}. \] Next, we need a common denominator to combine these fractions. The common denominator for \(4\) and \(2\) is \(4\): - Convert \(3\) to a fraction with a denominator of \(4\): \[ 3 = \frac{12}{4}. \] - Convert \(-\frac{3}{2}\) to a fraction with a denominator of \(4\): \[ -\frac{3}{2} = -\frac{6}{4}. \] Now we can rewrite the expression: \[ \frac{12}{4} - \frac{3}{4} - \frac{6}{4} + \frac{1}{4}. \] Combine the fractions: \[ \frac{12 - 3 - 6 + 1}{4} = \frac{4}{4} = 1. \] Thus, the final result is \[ \boxed{1}. \]