\( \tan ^ { 2 } 30 ^ { \circ } - \operatorname { Cos } 0 ^ { \circ } + L 5 \operatorname { Sin } ^ { 2 } 60 \)

We begin with the expression \[ \tan^2 30^\circ - \cos 0^\circ + L5\,\sin^2 60^\circ. \] It is common to interpret \( L5 \) as the natural logarithm \(\ln 5\). With that in mind, we proceed as follows. 1. **Evaluate \(\tan^2 30^\circ\):** We know that \[ \tan 30^\circ = \frac{1}{\sqrt{3}}. \] Squaring gives \[ \tan^2 30^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}. \] 2. **Evaluate \(\cos 0^\circ\):** \[ \cos 0^\circ = 1. \] 3. **Evaluate \(\sin^2 60^\circ\):** Since \[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \] squaring yields \[ \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}. \] 4. **Substitute the values into the expression:** Replacing each term, we have \[ \frac{1}{3} - 1 + \ln 5 \cdot \frac{3}{4}. \] 5. **Combine the terms:** Rewrite the constant terms with a common denominator: \[ \frac{1}{3} - 1 = \frac{1}{3} - \frac{3}{3} = -\frac{2}{3}. \] Thus, the expression becomes \[ -\frac{2}{3} + \frac{3\ln5}{4}. \] The final answer is \[ \frac{3\ln 5}{4} - \frac{2}{3}. \]