The given expression to evaluate is:

$\frac{2 \sin^2 60^\circ - \tan^2 30^\circ}{\sec^2 45^\circ}$

Step-by-step solution:

1. Evaluate the trigonometric values:

   • $\sin 60^\circ = \frac{\sqrt{3}}{2}$
   • $\tan 30^\circ = \frac{1}{\sqrt{3}}$
   • $\sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$

2. Square the trigonometric values:

   • $\sin^2 60^\circ = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}$
   • $\tan^2 30^\circ = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3}$
   • $\sec^2 45^\circ = (\sqrt{2})^2 = 2$

3. Substitute these values into the expression: $\frac{2 \left( \frac{3}{4} \right) - \frac{1}{3}}{2}$

4. Simplify the numerator: $2 \times \frac{3}{4} = \frac{6}{4} = \frac{3}{2}$ So the numerator becomes: $\frac{3}{2} - \frac{1}{3}$

5. Find the common denominator for the numerator: $\frac{3}{2} = \frac{9}{6}, \quad \frac{1}{3} = \frac{2}{6}$ Now, subtract: $\frac{9}{6} - \frac{2}{6} = \frac{7}{6}$

6. Final expression: $\frac{\frac{7}{6}}{2} = \frac{7}{6} \times \frac{1}{2} = \frac{7}{12}$

Thus, the value of the expression is:

$\boxed{\frac{7}{12}}$

Let me know if you would like further details or clarifications!

Related Questions:

1. What is the general formula for $\sin^2 \theta$ and $\tan^2 \theta$?
2. How does the identity $\sec^2 \theta = 1 + \tan^2 \theta$ apply to this problem?
3. How can we use the Pythagorean identity to simplify trigonometric expressions?
4. What are the key values of $\sin$, $\tan$, and $\sec$ for standard angles (e.g., $30^\circ$, $45^\circ$, $60^\circ$)?
5. Can you express the final result $\frac{7}{12}$ as a decimal?

Tip: Always check if common trigonometric identities can simplify the problem further!